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Tensile Trading: The 10 Essential Stages of Stock Market Mastery
Tensile Trading: The 10 Essential Stages of Stock Market Mastery
- Set yourself up for success by implementing prudent money management and asset protection strategies
- Build a personalized Asset Allocation Profile—your own personalized investment methodology
- Construct a properly diversified portfolio using tools and techniques tailored for the modern market
- Learn to take control of your "Investor Self," limiting the impact of mental hurdles and emotional baggage
- Supercharge your financial analysis by employing proven routines and strategies
A clear and proven approach easily tailored to fit your specific investing style, Tensile Trading distills the vastness of the financial markets into ten essential stages. It is designed to provide a comprehensive structure to your financial management efforts—helping you make smarter investment decisions, trade more efficiently, and consistently earn greater returns.
Editorial Reviews
- John Murphy, author of Technical Analysis of the Financial Markets and Trading with Intermarket Analysis; Chief Technical Analyst, StockCharts.com; former technical analyst, CNBC
- Dr. Henry Pruden, Professor of Business Administration and Executive Director, Institute of Technical Market Analysis, Golden Gate University; former editor, Market Technicians Association Journal
- Thom Hartle, Director of Product Trading, CQG, Inc.; former editor, Stocks & Commodities Magazine
Roger Shantz, individual investor; former mechanical engineer and naval aviator
About Gatis Roze:
A veteran full-time stock market investor who has traded his own account since 1989 unburdened by the distraction of clients. He holds an MBA from the Stanford Graduate School of Business, is a Chartered Market Technician (CMT) and is a past president of the Technical Securities Analysts Association (TSAA). After several successful entrepreneurial ventures, he retired in his early 40s to focus on investing in the financial markets. With consistent success as a stock market trader, he began teaching investments at the post-college level in 2000 and continues to do so today. A renowned communicator, he has taught thousands of investors in sold-out seminars for the American Association of Individual Investors (AAII), the Market Technicians Association (MTA), Bellevue College and StockCharts University. He also writes the popular blog, The Traders Journal, which has a global following and appears weekly at StockCharts.com.
About Grayson Roze:
Has worked in the financial services industry for StockCharts.com since 2012. He now serves as a Business Manager at the company. Grayson holds a Bachelor's degree from Swarthmore College, where he studied Economics and Psychology. At a young age, he began pursuing his interest in the financial markets by attending investment classes and starting to trade under the guidance of his father. At the age of 18, Grayson began investing his own account and has since become an accomplished trader in his own right. He speaks regularly at various investment seminars, including to organizations such as the American Association of Individual Investors (AAII) and Bellevue College.
Hardcover: 208 pages
Publisher: Wiley; 1st Edition (April 2016)
Availability: In Stock - Usually ships the next business day.
Browse more Book or Gatis Roze and Grayson Roze products.


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Join leading Blockchain CTOs and developers from ConsenSys, Eris, Coinbase, BlockStack, BlockCypher, Coinigy and more to explore Blockchain tech much deeper at Blockchain With The Best! Our pristine speakers will cover all Blockchain tech themes in 30min talks followed by interactive Q&A. Do you have a speaker that you must talk to? Then pre-book a 1-to-1 session and have your own time with a Blockchain guru.
Talks are accessible on Replay available for 2 months after the event to attendees.
We are waiting for you Saturday, June 11th from 10am / 6pm (EDT) on Blockchain.withthebest.com
Twitter @LearnWTB #BlockchainWTB

subset: commoncrawl


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The winless Miami Dolphins travel to San Diego this week, not that the locals will see Reggie Bush's homecoming on live TV.
The game will be blacked out locally after the Chargers failed to sell it out. If Qualcomm Stadium continues to be partially empty, talk of a prospective move to Los Angeles in the near future will likely be further fueled.
More than 6,500 tickets remained available Thursday, so many that the club didn't seek an extension before announcing the blackout.
Nate Davis is a reporter, blogger and editor who's been at USA TODAY since 2000. He has covered the NFL since 2005. No, he did not play quarterback for Ball State. Davis' succession of our esteemed colleague Sean Leahy at The Huddle is considered a Brady-for-Bledsoe swap by most "insiders."More about Nate


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What is JotForm?
JotForm is a free online form builder which helps you create online forms without writing a single line of code. No sign-up required.
At JotForm, we want to make sure that you're getting the online form builder help that you need. Our friendly customer support team is available 24/7.
We believe that if one user has a question, there could be more users who may have the same question. This is why many of our support forum threads are public and available to be searched and viewed. If you'd like help immediately, feel free to search for a similar question, or submit your question or concern.
How can I make my current account a non profit account?Asked by oia_acs on December 28, 2016 at 01:42 PM
I am signed up as a regular user but I want to upgrade to a paid account using the non-profit discount
JotForm Support
You can apply for non profit discount from your current account.
Use this link to apply for discount:
Hope this will help. Let us know if you need further assistance.


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Advertisement [CLICK THE BUTTON ON THE RIGHT TO CONTINUE]
Author | Tasneem Limbdiwala | Wednesday, Feb 28,2007 8:02 AM
Nike India has unveiled its latest ad campaign 'Nike Mean Street' and the creative, that has many talking at present, is every bit as superior in quality as some of the international Nike advertising. JWT Bangalore is the agency and the creative brain behind the campaign is Agnello Dias. Abhinav Deo has produced the film.
Even as the campaign is released just prior to Cricket World Cup 2007 and will be on-air for the next two months, across television, internet and cinema and Nike is a sponsor too, Sanjay Gangopadhyay, Director-Marketing, Nike India, clarifies that the campaign is not as much about the World Cup, as it is about cricket per se.
According to Gangopadhyay, the marketing programmes are not uni-dimensional. He said, "There are different ways in which Nike will be manifesting its relationship with cricket. From the timing perspective, cricket fever is on the peak and we are talking cricket. However, the ad has not been created to say it is an ad around the World Cup '07. Thus we are not saying 'go India, do the country proud'."
Dhunji Wadia, Senior VP & GM, JWT Bangalore, seconded this view stating that the film is about "pure, unadulterated cricket". He elaborated, "The creative execution features cricket crazy 15-17-year olds who play a tough game of cricket giving us scope to make the action really dramatic. Advertising and other imagery regarding cricket in India has revolved around either the cricket stars or really sweet kids playing street cricket. However no one has truly managed to capture the way the game is really played on the streets of India," said Wadia.
The campaign is around boys actually playing a hard and tough game who can swing a mean ball or drive a bludgeoning bat on the 'maidans' and gullies of India. "And these are guys really, truly, unabashedly representing the passion for the game in India" remarked Wadia.
The TVC also features Sreesanth and Zaheer Khan along with under-19 players Ryan Ninan and Uday Kaul who are Nike athletes.


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Profound Personal Integrity
"From a humble admission of self deception integrity is born."
We are going to visit the arena of Profound Humanness called "Integrity". Sometimes "integrity" is reduced to mean a kind of moral uprightness and steadfastness, in the sense of saying, "You have too much integrity to ever take a bribe".
But profound integrity goes far beyond this. Sometimes, in order to distinguish it from more limited popular usage, it is called "secondary integrity". This is the integrity which is not constrained by limited moralities, however well intentioned. The integrity that is profound living is the singularity of thrust of a life committed and ordering every dimension of the self towards that commitment. Thus the self is in fact shaped by the self, and focused towards that commitment. You can say that an audacious creation of the self takes place in integrity, without which you are simply the creation of the various forces impacting you in your society.
Thus the basis of integrity is a destinal resolve – a resolve that chooses and sets your destiny and out of which your whole life is ordered. The object of that resolve is the ultimate decision of each person, and each person makes that choice, consciously or unconsciously. To do so with awareness is the height of our responsibility. It is incarnate freedom. It is what real freedom looks like. When you have thus exercised your freedom you realize that to be true to yourself ever thereafter you have a unique position to look at the values of society. You are no longer bound by the opinions and codes of others, but reevaluate then on the basis of their impact on your destinal resolve.
Thus the person of integrity is continuously engaged in a societal transvaluation, a moving across the values of society and reinterpreting them in line with their life's thrust. It does not give you the liberty of ignoring your society, but your obligation transcends the conformity of living within the codes and mores of your society. Thus the person of profound integrity always seems to not quite fit with everyone else, but your actions always are appropriate for you, even to those who oppose you.
No matter how odd the person of profound integrity appears to their neighbors, they experience themselves as securely anchored. While you are very clear that this world is not your home, nevertheless you experience yourself as having found your native vale. You experience an eternal at-one-ness, not so much with the currents and waves of activity around you, but with the deeper trends of history itself. Amid the flux of wavering to and fro that is so evident in others, you experience an inexplicable rootedness, as though you have sunk a taproot deep into the foundations of the earth itself. Though you experience life as a long journey, even an endless journey, towards the object of your resolve, yet you never sense yourself as a stranger on the journey. It's as if you'd been there before. Original integrity is experienced primarily by this sense of at-one-ness.
Kierkegaard once wrote a book about this kind of integrity that he titled, "Purity of Heart is to Will One Thing". An ancient philosopher focused his wisdom around this integrity with the advice, "Know yourself, and to your own self, be true".
"This document comes from the curriculum of the Maliwada Human Development Training School in Maliwada, India, which is a former program of ICA International They are not certain of the year, so there are no further details that they could provide for a citation." I have rewritten this article by taking out the reference to "man" or "he", and replacing it with "you" or "yourself" so that women will not feel left out when reading the text and so that it feels more personal rather than third-party.
Above quoted from an email dated 13 Feb. 2008 sent to me by the Director of Development and Communications at The Institute of Cultural Affairs International (ICA) in Quebec, Canada.
I have rewritten this article by taking out the reference to "man" or "he", and replacing it with "you" or "yourself" so that women will not feel left out when reading the text and so that it feels more personal rather than third-party.
P.S. Plus, here is another of my favorites... Desiderata


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Yesterday we showed you how to find best selling items on eBay by using the completed listings search. Today we're going to look at another way to find hot markets: the eBay Pulse.
The eBay Pulse gives you a snapshot of what's hot in each product category. You can use the tool to:
You can find eBay Pulse here: http://pulse.ebay.com/. Or to navigate there from within eBay, just mouseover on the "Sell" menu item across the top. Click on "What's Hot" form the drop-down list. This brings you to the main Pulse page.
We highly recommend viewing the Pulse daily and putting your findings into a spreadsheet.
Discover four of the most profitable niches on eBay and get details for the best suppliers for these niches. Enter your details below and we will send them instantly.
Your list of profitable markets and supplier details are on their way to your email inbox.
Be sure to add firstname.lastname@example.com to your safelist to ensure we can deliver the free suppliers details to you.
If you haven't received an email from us in 5 minutes, please check your spam folder or email us at firstname.lastname@example.com.


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Cedarbrook Lodge, located near Sea-Tac International Airport, boasts over 18,000 square feet of interior event space and 5,500 square feet of diverse outdoor venues, making it a meeting facility ideally suited for events from 8 to 300.
Here at Cedarbrook Lodge we pride ourselves on our attention to detail. Please fill out the following form to submit a proposal for a memorable Seattle-area meeting or event. You may expect to be contacted by one of our sales associates within one business day. We look forward to working with you soon!
>> Contact Information
>> Meeting or Event Information
>> Room Requirements
>> Meeting & Event Details


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Across the four east London estates collectively known as the Parkside Estate, PRP has developed a number of small infill sites to address current overcrowding in the area. These sites are to be occupied by high quality family homes with mixed tenure, and range in size from single-storey conversions to a large site of over 40 new homes. The primary objective was to create a sense of place that fully integrates with the surrounding environments and is unique to each site. The sites have been controversial and PRP has worked closely with the community to address any concerns.
PRP and Circle Housing Old Ford's regeneration of Hitchin Square has transformed a dilapidated garage site and underused grass area into a new mews street with landscaped area for play, relaxation and visual amenity.
The landscape design helps to bind the new and existing building styles and reactivates the communal open space through providing new paths and activities. The layout for this area is a simple oval of grass around which play, planting and a footpath are arranged. The existing trees were retained to provide a backdrop to the new buildings and because they provide a sense of scale and maturity to the space.
Planning at PRP worked in close collaboration with our architects and Circle Housing Group to regenerate and redevelop 18 sites across the Parkside Estate in the London Borough of Tower Hamlets. Owing to the constrained nature of many of the sites, innovative proposals were developed to ensure that high quality affordable dwellings were proposed that responded to the environment and complemented the existing dwellings and communities within the Estate. The last site, a brownfield redevelopment for 45 affordable homes adjacent to the A12 is currently being determined by the Council.


text: I know there an option there when creating forums (nodes), to add pages and was wondering how easy it is to create a page used for the homepage link pulling some basic information from the forum onto it. Much like a Portal really and it doesn't have to be complicated with lots of features. Just something basic like pulling latest threads and replies on it would do just fine. Are there any good guides how to do this fast, as I really do "NOT" understand the working of XenForo code in general to do it myself. I'm aware there's a Portal add-on by JAX before anyone points me in that direction, but it's not what I'm looking for here. I'm trying to stay away from using add-ons keeping my board running stock.


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Las Cruces, also known as "The City of the Crosses", is the county seat of Doña Ana County, New Mexico, United States. The population was 97,618 in 2010 according to the 2010 Census, making it the second largest city in the state. It is the principal city of a Metropolitan Statistical Area which encompasses all of Dona Ana County and is part of the larger El Paso-Las Cruces, TX-NM Combined Statistical Area.
- Nitro - August 28, 2000.
- WWE House Show - January 22, 2012.
External links
- Las Cruces, New Mexico on Wikipedia


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Model: 05013
Keep all of your maintenance/cleaning equipment safe and out of the way with the MonkeyBar Mop & Broom Storage Rack! The simple yet effective combination of hooks maximize its storage potential making this rack nearly limitless. Order yours today!
• 5 different styles of hooks
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Click OK to view our full shipping policy. Clicking cancel will keep you on the current page.


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Parenting by many measures is certainly a full-time job in its own right. And if you're a working parent, you probably know how challenging it can be to raise a child and work a full-time job at the same time. And often, some parents decide to leave their work entirely. If that is not an option for you, you might consider to look for some career opportunities with decent earnings and flexible hours. Check out our list of some of the best flexible jobs for working parents.
Account Manager
Account managers are employed across a wide variety of industries including advertising, financial services and media. Advertising and promotions managers bring in about ₦150,000 a month. While you may have to deal with client relationships, you might be able to perform most of your duties off-site. Being able to send emails and join conference calls from the comfort of your own home can be particularly helpful to parents with young children.
Here are hot job openings in accounting management.
Web Designer
If you have a great eye for design, and lots of experience with computers—possibly also programming languages and a bachelor's degree, then you can make great money, set your own hours and projects, and do web design as your career. It's possible to find an employer who can provide you with a schedule that best fits your needs.
Check out the list of hot web-design vacancies on job.naij.com
A dietitian advises clients what to eat or serve in order to achieve a specific health goal or to simply maintain a healthy lifestyle. A dietitian may work in a healthcare facility, school, or other organization.
Dietitians or nutritionist can also be self-employed - an option for parents who want to create their own, flexible schedules.
The median salary for nutritionists is from ₦80 000 to ₦150 000. Since the projected growth rate is better than average, becoming a registered dietitian might be worth it.
Here are some opportunities you might want to consider.
Software Developer/Coder
This kind of job can be intimidating to folks who don't know how to code. But it's never too late to learn. In fact, there are all sorts of free classes online that can teach you the basics. Coding is considered a must-have skill, so if you're a coder, it shouldn't be too difficult to find a high-paying gig that you can jump into whether the kids are at school or at home with you.
While you may need to have some experience or advanced credentials, being a healthcare, financial or marketing consultant can bring in some serious cash. Plus, these jobs often offer a great deal of flexibility. To get started, you can find clients within your existing network and use those connections to find other people to advise later down the line.
Do you know another language? Are you willing to learn one? You'll probably need a bachelor's degree, plus superb communication skills—not just in English, but in multiple languages. But if this is your skill set, you can make your own hours and design a schedule that works best for you. The need for multinational communication continues to grow and as an interpreter or translator, you might be able to work from home and take on projects as needed. Regardless of which direction you head in this field, you will likely make your own hours and retain optimal flexibility that is so important when you have children.
Personal Trainer
Many parents leave the workforce for a time, but some never stop hitting the gym. Maybe it's time to turn all of that extra effort into a living.
If you haven't taken a break from the gym the way you have from the workforce and you can pass the necessary licensing tests, then you can make a good amount of money helping other people to get fit. Bonus: you'll get or stay fit yourself.
Whether you want to be a teacher, an aide or a curriculum developer, a job in education can be a great fit for an individual trying to raise a family, especially since you'll likely have more vacation time and work hours that won't prevent you from caring for your children. You may have to get a master's degree, however, if you want a shot at a competitive salary.
Check out the list of hot vacancies in education on job.naij.com
Event Planner
With all the skills you have planning birthday parties and family cookouts, this is just the big league equivalent.
If you're the type of person who excels in a fast-paced, detail-oriented atmosphere, then event planning could be the career for you.
Just as it is with parenting, you'll have to coordinate everything seamlessly and be able to think on your feet to make sure any problems that do arise are taken care of quickly. If you're a born networker who can maintain multiple relationships with vendors and routinely pull off gatherings without a hitch, this is a great job for you.
Freelance Writer
If you're looking for a job that harnesses your creativity, revolves around interesting topics and allows maximum freedom and flexibility, consider becoming a web writer.
This job can take on many forms. Find an industry in which you have an interest and experience, and then see if any companies need someone to provide content. You'll have to adhere to industry standards and strive to always meet company guidelines regarding your content..
Or you could do what many parents have done in the last few years and become a blogger. Find a subject about which you're passionate and market it to the point you can sell advertising. Not everyone can be a Dooce or Bloggess, but the opportunity is there.
Choosing a job can be difficult, especially when you have kids to take care of. As you hunt for opportunities, it's a good idea to think about your projected job growth and make sure you have enough time to be the kind of parent you want to be.


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Blossom at Wikia
Welcome to the Blossom mini wiki at Scratchpad!
You can use the box below to create new pages for this mini-wiki. Make sure you type
[[Category:Blossom]] on the page before you save it to make it part of the Blossom wiki (preload can be enabled to automate this task, by clicking this link and saving that page. Afterwards, you may need to purge this page, if you still see this message).
Creative Ad Agency
Captions, Logos, Tag Lines, Business Philosophy and Concept creation.


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Fayette County woman goes from gas fields to silver screen
April Black, 24, of Smithfield in Fayette County, works in the Marcellus Shale gas-drilling industry.
That's her day job. She's also an actress in "The Hunger Games" -- as well as movies like "Abduction," "Super 8," and this summer's much-anticipated "The Avengers." She became a stuntwoman for the film "Freerunner," performing whenever things got too hot for the Brazilian supermodel playing the lead role. She also works as a model for American Eagle and runway shows at New York City's Fashion Week, among others.
Her story sounds like it could be a sequel to "Flashdance." But Black sees her recently acquired day job in the gas fields of Western Pennsylvania more as a source of stability -- which can be in short supply at the working-class level of the acting and modeling business.
"It's a complete 180," she says. "I have a 40-hour week, doing a lot of construction projects, dealing with some billing for construction projects, scheduling meetings, things like that. I've got to take care of bills, and want to get a house and take care of retirement, and save for the long run. Sometimes (acting's) not really reliable, but it's really fun and great traveling and doing all that stuff."
Her featured role in "The Hunger Games" was a particularly big deal, since she's a huge fan of Suzanne Collins' books.
"It's funny, because I started reading the books maybe three years before I heard anything about the movie," Black says. "I told all my friends about them, and said 'If they ever make a movie about these books, I don't care what I have to do to be there, I'll be there.'"
She went to North Carolina for casting, and was overwhelmed by the number of people clamoring to be in the movie.
"They said people showed up at the casting from 49 out of 50 states of our country, and people from 20 other countries around the world," says Black. "It got so big so fast."
Oddly enough, another person who made the cut was her friend Dayna Kutek, 16, of McClellandtown in Fayette County.
Black wasn't allowed to divulge too much about her role, but says that she played a woman from the Capitol -- the wealthy enclave that pits children from vassal states in duels to the death for their televised entertainment.
She got to act in scenes with Donald Sutherland, Woody Harrelson and Stanley Tucci.
One day on the set, she was surprised to find herself sitting next to Lenny Kravitz.
"It was really exciting, because nobody really knew he was there," she says. "He's a really humble person, which is surprising for someone who has achieved so much fame. He sat next to me, and said, 'Hey, I'm Lenny. What's your name?' He was such a cool person. Some people get all nervous. They tell you when you go in to shoot some scenes that you're not allowed to ask for autographs, because then they get real uncomfortable. You can see why -- they just want to be treated like a normal person. He didn't want any special treatment."
One scene was a particular challenge: How does one react to watching children kill each other in an arena• What if you have money riding on it?
"One of the things was a featured close-up of me reacting to the murders that are going on. The 'tributes' are trying to survive in the arena -- kill each other to advance to the next level. The children are ages 12 to 18. (The director) told me, 'You're looking at this person killing this person, but you're winning money because you're betting on them.' It was challenging. It's a bit heart-wrenching. How am I able to make this come across -- that I'm loving what I'm seeing?"


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"Dipping Your Pen In The Company Ink" Is More Common Than You Think
Have you thought about having an office affair with someone? Read this before you do.
Have you thought about having an office affair with someone? Read this before you do.
Forrest Gump once said life is like a box of chocolates. But that was way back in the '90s. Nowadays, life is like an email.
These jokes are lame, but they are all in good pun.
When it comes to working in an office, you work with a bunch of personalities. Sometimes you have to allow for the occasional disagreement, however sometimes, things just really irritate you. There are so many things that other people do that annoy another, and vice versa. So what are some work place pet peeves. I polled some people around my office.
When you're working in close quarters with the same people for long periods of time, the air can get a bit hostile now and then. While there's no way everyone in an office can get along perfectly, there are a few employees who find it hard to tolerate offensive actions, like eating other peoples' lunches or stuffing toilet paper in between the stalls so no one sees them on the toilet. It happens.
As Lori had mentioned in an earlier story, today is 'Love your pet day.'
In her Top 5 list of ways to celebrate, I was surprised she didn't put 'bring your pet to the office.'
An eagle-eyed tipster has sent the New York Post a screenshot from last week's episode of 'The Office" which seems to show Steve Carell, the show's former star, making an unannounced cameo.
The communal offices that were introduced during the '50s to boast creativity and problem solving are actually reducing productivity, according to new research.
Neuroscientist Dr. Jack Lewis conducted brain scans on office workers and found that those who work in spaces with few walls are plagued by intense bursts of distraction.


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Best foam parties in WA!
"The most awesome fun party we've ever had! FoamZone rocks!" - Amy
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What is a foam party?
A foam party is the best thing!
The idea of dancing away and having thousands of litres of foam sprayed and poured all over you until you are covered to neck height may sound a bit radical, but the reality is that it is the greatest and most enjoyable, fun thing to be added to a party.
Brace yourself for a wild party!
Foam parties are immensely popular all over the world. Many big clubs in London, Los Angeles, New York etc have made huge successes from foam party nights and now, Foamzone has brought Perth to the scene as well.
A foam party is a great way to promote a club or any event.
It's best to see for yourself what a foam party looks like, just click here.
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Are foam parties safe?
Foamzone foam parties are very safe, we only use foam that is safe, non-toxic and environmentally friendly. The foam is low-irritant to skin and eyes and won't damage clothes or floors. It is not slippery like soap. It has the same viscosity as water and does not have soap-like qualities.


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Fellow nerds, prepare for a most excellent colliding of two worlds. Dollhouse, the anticipated new drama from Buffy creator Joss Whedon, has added a familiar face from another sci-fi galaxy not so far away: Battlestar Galactica.
Starring opposite Eliza Dushku will be Tahmoh Penikett, who plays Karl "Helo" Agathon in the Sci-Fi Channel's critically praised drama. Penikett's character in Dollhouse is an FBI agent obsessed with learning about the Dollhouse, a secret organization that imprints memories and personalities into people to serve missions. Penikett and Dushku will also become romantically entangled on the show.
Also added to the cast was Fran Kranz, who recently headed up Welcome to the Captain. Kranz's character will be a programmer in charge of installing the personalities into the "dolls." With Kranz on board, it looks as if the poorly received Welcome to the Captain is pretty much a lock for cancellation.
Neighbours's hapa actress Dichen Lachman and The Unit guest star Enver Gjokaj have also been cast.


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Number Theory for Mathematical Contests. David A. SANTOS
- Jacob McCoy
- 2 years ago
- Views:
1 Number Theory for Mathematical Contests David A. SANTOS August 13, 2005 REVISION
2 Contents Preface 1 Preliminaries Introduction Well-Ordering Practice Mathematical Induction Practice Fibonacci Numbers Practice Pigeonhole Principle Practice Divisibility Divisibility Practice Division Algorithm Practice Some Algebraic Identities Practice Congruences. Z n Congruences Practice Divisibility Tests Practice Complete Residues Practice Unique Factorisation GCD and LCM Practice Primes Practice Fundamental Theorem of Arithmetic Practice iii 5 Linear Diophantine Equations Euclidean Algorithm Practice Linear Congruences Practice A theorem of Frobenius Practice Chinese Remainder Theorem Practice Number-Theoretic Functions Greatest Integer Function Practice De Polignac s Formula Practice Complementary Sequences Practice Arithmetic Functions Practice Euler s Function. Reduced Residues Practice Multiplication in Z n Practice Möbius Function Practice More on Congruences Theorems of Fermat and Wilson Practice Euler s Theorem Practice Scales of Notation The Decimal Scale Practice Non-decimal Scales Practice A theorem of Kummer Miscellaneous Problems 91 Practice
3 Preface These notes started in the summer of 1993 when I was teaching Number Theory at the Center for Talented Youth Summer Program at the Johns Hopkins University. The pupils were between 13 and 16 years of age. The purpose of the course was to familiarise the pupils with contest-type problem solving. Thus the majority of the problems are taken from well-known competitions: AHSME AIME USAMO IMO ITT MMPC (UM) 2 STANFORD MANDELBROT American High School Mathematics Examination American Invitational Mathematics Examination United States Mathematical Olympiad International Mathematical Olympiad International Tournament of Towns Michigan Mathematics Prize Competition University of Michigan Mathematics Competition Stanford Mathematics Competition Mandelbrot Competition Firstly, I would like to thank the pioneers in that course: Samuel Chong, Nikhil Garg, Matthew Harris, Ryan Hoegg, Masha Sapper, Andrew Trister, Nathaniel Wise and Andrew Wong. I would also like to thank the victims of the summer 1994: Karen Acquista, Howard Bernstein, Geoffrey Cook, Hobart Lee, Nathan Lutchansky, David Ripley, Eduardo Rozo, and Victor Yang. I would like to thank Eric Friedman for helping me with the typing, and Carlos Murillo for proofreading the notes. Due to time constraints, these notes are rather sketchy. Most of the motivation was done in the classroom, in the notes I presented a rather terse account of the solutions. I hope some day to be able to give more coherence to these notes. No theme requires the knowledge of Calculus here, but some of the solutions given use it here and there. The reader not knowing Calculus can skip these problems. Since the material is geared to High School students (talented ones, though) I assume very little mathematical knowledge beyond Algebra and Trigonometry. Here and there some of the problems might use certain properties of the complex numbers. A note on the topic selection. I tried to cover most Number Theory that is useful in contests. I also wrote notes (which I have not transcribed) dealing with primitive roots, quadratic reciprocity, diophantine equations, and the geometry of numbers. I shall finish writing them when laziness leaves my weary soul. I would be very glad to hear any comments, and please forward me any corrections or remarks on the material herein. David A. SANTOS iii
4 Legal Notice This material may be distributed only subject to the terms and conditions set forth in the Open Publication License, version 1.0 or later (the latest version is presently available at THIS WORK IS LICENSED AND PROVIDED AS IS WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IM- PLIED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE OR A WARRANTY OF NON-INFRINGEMENT. THIS DOCUMENT MAY NOT BE SOLD FOR PROFIT OR INCORPORATED INTO COMMERCIAL DOCUMENTS WITHOUT EXPRESS PERMISSION FROM THE AUTHOR(S). THIS DOCUMENT MAY BE FREELY DISTRIBUTED PROVIDED THE NAME OF THE ORIGINAL AUTHOR(S) IS(ARE) KEPT AND ANY CHANGES TO IT NOTED. iv
5 Chapter1 Preliminaries 1.1 Introduction We can say that no history of mankind would ever be complete without a history of Mathematics. For ages numbers have fascinated Man, who has been drawn to them either for their utility at solving practical problems (like those of measuring, counting sheep, etc.) or as a fountain of solace. Number Theory is one of the oldest and most beautiful branches of Mathematics. It abounds in problems that yet simple to state, are very hard to solve. Some number-theoretic problems that are yet unsolved are: 1. (Goldbach s Conjecture) Is every even integer greater than 2 the sum of distinct primes? 2. (Twin Prime Problem) Are there infinitely many primes p such that p+2 is also a prime? 3. Are there infinitely many primes that are 1 more than the square of an integer? 4. Is there always a prime between two consecutive squares of integers? In this chapter we cover some preliminary tools we need before embarking into the core of Number Theory. 1.2 Well-Ordering The set N = {0,1,2,3,4,...} of natural numbers is endowed with two operations, addition and multiplication, that satisfy the following properties for natural numbers a,b, and c: 1. Closure: a+b and ab are also natural numbers. 2. Associative laws: (a+b)+c = a+(b+c) and a(bc) = (ab)c. 3. Distributive law: a(b+c) = ab+ac. 4. Additive Identity: 0+a = a+0 = a 5. Multiplicative Identity: 1a = a1 = a. One further property of the natural numbers is the following. 1 Axiom (Well-Ordering Axiom) Every non-empty subset S of the natural numbers has a least element. As an example of the use of the Well-Ordering Axiom, let us prove that there is no integer between 0 and 1. 2 Example Prove that there is no integer in the interval ]0;1[. 1
6 2 Chapter 1 Solution: Assume to the contrary that the set S of integers in ]0;1[ is non-empty. Being a set of positive integers, it must contain a least element, say m. Now, 0 < m 2 < m < 1, and so m 2 S. But this is saying that S has a positive integer m 2 which is smaller than its least positive integer m. This is a contradiction and so S =. We denote the set of all integers by Z, i.e., Z = {... 3, 2, 1,0,1,2,3,...}. A rational number is a number which can be expressed as the ratio a of two integers a,b, where b 0. We denote the set of b rational numbers by Q. An irrational number is a number which cannot be expressed as the ratio of two integers. Let us give an example of an irrational number. 3 Example Prove that 2 is irrational. Solution: The proof is by contradiction. Suppose that 2 were rational, i.e., that 2 = a for some integers a,b. This implies b that the set A = {n 2 : both n and n 2 positive integers} is nonempty since it contains a. By Well-Ordering A has a smallest element, say j = k 2. As 2 1 > 0, j( 2 1) = j 2 k 2 = ( j k) 2 is a positive integer. Since 2 < 2 2 implies 2 2 < 2 and also j 2 = 2k, we see that ( j k) 2 = k(2 2) < k( 2) = j. Thus ( j k) 2 is a positive integer in A which is smaller than j. This contradicts the choice of j as the smallest integer in A and hence, finishes the proof. 4 Example Let a,b,c be integers such that a 6 + 2b 6 = 4c 6. Show that a = b = c = 0. Solution: Clearly we can restrict ourselves to nonnegative numbers. Choose a triplet of nonnegative integers a, b, c satisfying this equation and with max(a,b,c) > 0 as small as possible. If a 6 + 2b 6 = 4c 6 then a must be even, a = 2a 1. This leads to 32a b6 = 2c 6. Hence b = 2b 1 and so 16a b6 1 = c6. This gives c = 2c 1, and so a b6 1 = 4c6 1. But clearly max(a 1,b 1,c 1 ) < max(a,b,c). This means that all of these must be zero. 5 Example (IMO 1988) If a,b are positive integers such that a2 + b 2 1+ab is an integer, then a2 + b 2 1+ab is a perfect square. Solution: Suppose that a2 + b 2 = k is a counterexample of an integer which is not a perfect square, with max(a,b) as small as 1+ab possible. We may assume without loss of generality that a < b for if a = b then 0 < k = 2a2 a < 2, which forces k = 1, a perfect square. Now, a 2 + b 2 k(ab+1) = 0 is a quadratic in b with sum of the roots ka and product of the roots a 2 k. Let b 1,b be its roots, so b 1 + b = ka and b 1 b = a 2 k. As a,k are positive integers, supposing b 1 < 0 is incompatible with a 2 + b 2 1 = k(ab 1 + 1). As k is not a perfect square, supposing b 1 = 0 is incompatible with a = k(0 a+1). Also b 1 = a2 k b < b2 k b < b.
7 Practice 3 Thus we have found another positive integer b 1 for which a2 + b ab 1 = k and which is smaller than the smallest max(a,b). This is a contradiction. It must be the case, then, that k is a perfect square. Practice 6 Problem Find all integer solutions of a 3 + 2b 3 = 4c 3. 7 Problem Prove that the equality x 2 +y 2 +z 2 = 2xyz can hold for whole numbers x,y,z only when x = y = z = Mathematical Induction The Principle of Mathematical Induction is based on the following fairly intuitive observation. Suppose that we are to perform a task that involves a certain number of steps. Suppose that these steps must be followed in strict numerical order. Finally, suppose that we know how to perform the n-th task provided we have accomplished the n 1-th task. Thus if we are ever able to start the job (that is, if we have a base case), then we should be able to finish it (because starting with the base case we go to the next case, and then to the case following that, etc.). Thus in the Principle of Mathematical Induction, we try to verify that some assertion P(n) concerning natural numbers is true for some base case k 0 (usually k 0 = 1, but one of the examples below shows that we may take, say k 0 = 33.) Then we try to settle whether information on P(n 1) leads to favourable information on P(n). We will now derive the Principle of Mathematical Induction from the Well-Ordering Axiom. 8 Theorem (Principle of Mathematical Induction) If a sets of non-negative integers contains the integer 0, and also contains the integer n+1 whenever it contains the integer n, then S = N. Proof: Assume this is not the case and so, by the Well-Ordering Principle there exists a least positive integer k not in S. Observe that k > 0, since 0 S and there is no positive integer smaller than 0. As k 1 < k, we see that k 1 S. But by assumption k 1+1 is also in S, since the successor of each element in the set is also in the set. Hence k = k 1+1 is also in the set, a contradiction. Thus S = N. The following versions of the Principle of Mathematical Induction should now be obvious. 9 Corollary If a set A of positive integers contains the integer m and also contains n+1 whenever it contains n, where n > m, then A contains all the positive integers greater than or equal to m. 10 Corollary (Principle of Strong Mathematical Induction) If a set A of positive integers contains the integer m and also contains n+1 whenever it contains m+1,m+2,...,n, where n > m, then A contains all the positive integers greater than or equal to m. We shall now give some examples of the use of induction. 11 Example Prove that the expression is a multiple of 169 for all natural numbers n. 3 3n+3 26n 27 Solution: For n = 1 we are asserting that = 676 = is divisible by 169, which is evident. Assume the assertion is true for n 1,n > 1, i.e., assume that 3 3n 26n 1 = 169N for some integer N. Then 3 3n+3 26n 27 = n 26n 27 = 27(3 3n 26n 1)+676n
8 4 Chapter 1 which reduces to N n, which is divisible by 169. The assertion is thus established by induction. 12 Example Prove that is an even integer and that (1+ 2) 2n +(1 2) 2n (1+ 2) 2n (1 2) 2n = b 2 for some positive integer b, for all integers n 1. Solution: We proceed by induction on n. Let P(n) be the proposition: (1+ 2) 2n +(1 2) 2n is even and (1+ 2) 2n (1 2) 2n = b 2 for some b N. If n = 1, then we see that (1+ 2) 2 +(1 2) 2 = 6, an even integer, and (1+ 2) 2 (1 2) 2 = 4 2. Therefore P(1) is true. Assume that P(n 1) is true for n > 1, i.e., assume that (1+ 2) 2(n 1) +(1 2) 2(n 1) = 2N for some integer N and that (1+ 2) 2(n 1) (1 2) 2(n 1) = a 2 for some positive integer a. Consider now the quantity (1+ 2) 2n +(1 2) 2n = (1+ 2) 2 (1+ 2) 2n 2 +(1 2) 2 (1 2) 2n 2. This simplifies to (3+2 2)(1+ 2) 2n 2 +(3 2 2)(1 2) 2n 2. Using P(n 1), the above simplifies to 12N + 2 2a 2 = 2(6N + 2a), an even integer and similarly (1+ 2) 2n (1 2) 2n = 3a 2+2 2(2N) = (3a+4N) 2, and so P(n) is true. The assertion is thus established by induction. 13 Example Prove that if k is odd, then 2 n+2 divides k 2n 1 for all natural numbers n. Solution: The statement is evident for n = 1, as k 2 1 = (k 1)(k+ 1) is divisible by 8 for any odd natural number k because both (k 1) and (k + 1) are divisible by 2 and one of them is divisible by 4. Assume that 2 n+2 k 2n 1, and let us prove that 2 n+3 k 2n+1 1. As k 2n+1 1 = (k 2n 1)(k 2n + 1), we see that 2 n+2 divides (k 2n 1), so the problem reduces to proving that 2 (k 2n + 1). This is obviously true since k 2n odd makes k 2n + 1 even.
9 Mathematical Induction 5 14 Example (USAMO 1978) An integer n will be called good if we can write n = a 1 + a 2 + +a k, where a 1,a 2,...,a k are positive integers (not necessarily distinct) satisfying 1 a a a k = 1. Given the information that the integers 33 through 73 are good, prove that every integer 33 is good. Solution: We first prove that if n is good, then 2n+8 and 2n+9 are good. For assume that n = a 1 + a 2 + +a k, and Then 2n+8 = 2a 1 + 2a a k and Also, 2n+9 = 2a 1 + 2a a k and Therefore, 1 = 1 a a a k. 1 2a a a k = = a 1 2a 2 2a k = = 1. if n is good both 2n+8 and 2n+9 are good. (1.1) We now establish the truth of the assertion of the problem by induction on n. Let P(n) be the proposition all the integers n,n+1,n+2,...,2n+7 are good. By the statement of the problem, we see that P(33) is true. But (1.1) implies the truth of P(n+1) whenever P(n) is true. The assertion is thus proved by induction. We now present a variant of the Principle of Mathematical Induction used by Cauchy to prove the Arithmetic-Mean- Geometric Mean Inequality. It consists in proving a statement first for powers of 2 and then interpolating between powers of Theorem (Arithmetic-Mean-Geometric-Mean Inequality) Let a 1,a 2,...,a n be nonnegative real numbers. Then n a1 a 2 a n a 1 + a 2 + +a n. n Proof: Since the square of any real number is nonnegative, we have ( x 1 x 2 ) 2 0. Upon expanding, x 1 + x 2 x 1 x 2, 2 (1.2) which is the Arithmetic-Mean-Geometric-Mean Inequality for n = 2. Assume that the Arithmetic-Mean-Geometric- Mean Inequality holds true for n = 2 k 1,k > 2, that is, assume that nonnegative real numbers w 1,w 2,...,w 2 k 1 satisfy w 1 + w 2 + +w 2 k 1 2 k 1 (w 1 w 2 w 2 k 1) 1/2k 1. (1.3) Using (1.2) with x 1 = y 1 + y 2 + +y 2 k 1 2 k 1 and x 2 = y 2 k y 2 k 2 k 1,
10 6 Chapter 1 we obtain that y 1 + y 2 + +y 2 k 1 2 k 1 + y 2 k y 2 k 2 k 1 2 Applying (1.3) to both factors on the right hand side of the above, we obtain ( y 1 + y 2 + +y 2 k 1 2 k 1 )( y 2 k y 2 k 2 k 1 ) 1/2. y 1 + y 2 + +y 2 k 2 k (y 1 y 2 y 2 k) 1/2k. (1.4) This means that the 2 k 1 -th step implies the 2 k -th step, and so we have proved the Arithmetic-Mean-Geometric- Mean Inequality for powers of 2. Now, assume that 2 k 1 < n < 2 k. Let and Let Using (1.4) we obtain which is to say that y 1 = a 1,y 2 = a 2,...,y n = a n, y n+1 = y n+2 = = y 2 k = a 1 + a 2 + +a n. n A = a 1 + +a n n and G = (a 1 a n ) 1/n. a 1 + a 2 + +a n +(2 k n) a 1 + +a n n 2 k a 1 a 2 a n ( a 1 + +a n n) 1/2 k ) (2k, n This translates into A G or which is what we wanted. na+(2 k n)a 2 k (G n A 2k n ) 1/2k. (a 1 a 2 a n ) 1/n a 1 + a 2 + +a n, n 16 Example Let s be a positive integer. Prove that every interval [s;2s] contains a power of 2. Solution: If s is a power of 2, then there is nothing to prove. If s is not a power of 2 then it must lie between two consecutive powers of 2, i.e., there is an integer r for which 2 r < s < 2 r+1. This yields 2 r+1 < 2s. Hence s < 2 r+1 < 2s, which gives the required result. 17 Example Let M be a nonempty set of positive integers such that 4x and [ x] both belong to M whenever x does. Prove that M is the set of all natural numbers. Solution: We will prove this by induction. First we will prove that 1 belongs to the set, secondly we will prove that every power of 2 is in the set and finally we will prove that non-powers of 2 are also in the set. Since M is a nonempty set of positive integers, it has a least element, say a. By assumption a also belongs to M, but a < a unless a = 1. This means that 1 belongs to M. Since 1 belongs to M so does 4, since 4 belongs to M so does 4 4 = 4 2, etc.. In this way we obtain that all numbers of the form 4 n = 2 2n,n = 1,2,... belong to M. Thus all the powers of 2 raised to an even power belong to M. Since the square roots belong as well to M we get that all the powers of 2 raised to an odd power also belong to M. In conclusion, all powers of 2 belong to M.
11 Practice 7 Assume now that n N fails to belong to M. Observe that n cannot be a power of 2. Since n M we deduce that no integer in A 1 = [n 2,(n + 1) 2 ) belongs to M, because every member of y A 1 satisfies [ y] = n. Similarly no member z A 2 = [n 4,(n+1) 4 ) belongs to M since this would entail that z would belong to A 1, a contradiction. By induction we can show that no member in the interval A r = [n 2r,(n+1) 2r ) belongs to M. We will now show that eventually these intervals are so large that they contain a power of 2, thereby obtaining a contradiction to the hypothesis that no element of the A r belonged to M. The function f : R + R x log 2 x is increasing and hence log 2 (n+1) log 2 n > 0. Since the function f : R R + x 2 x is decreasing, for a sufficiently large positive integer k we have 2 k < log 2 (n+1) log 2 n. This implies that (n+1) 2k > 2n 2k. Thus the interval [n 2k,2n 2k ] is totally contained in [n 2k,(n+1) 2k ). But every interval of the form [s,2s] where s is a positive integer contains a power of 2. We have thus obtained the desired contradiction. Practice 18 Problem Prove that 11 n n+1 is divisible by 133 for all natural numbers n. 19 Problem Prove that equals 1 x 1! + x(x 1) 2! x(x 1)(x 2) 3! + +( 1) n x(x 1)(x 2) (x n+1) n! ( 1) n (x 1)(x 2) (x n) n! for all non-negative integers n. 20 Problem Let n N. ßÞ Prove the Ðinequality 1 n n n+1 > Problem Prove that Õ π 2 = 2cos 2 n+1 n radical signs for n N. 22 Problem Let a 1 = 3,b 1 = 4, and a n = 3 a n 1,b n = 4 b n 1 when n > 1. Prove that a 1000 > b Problem Let n N,n > 1. Prove that (2n 1) (2n) < 1 3n Problem Prove that if n is a natural number, then n (3n 1) = n 2 (n+1). 25 Problem Prove that if n is a natural number, then (2n 1) 2 = n(4n2 1) Problem Prove that for all natural numbers n > 1. 4 n n+1 < (2n)! (n!) 2 27 Problem Prove that the sum of the cubes of three consecutive positive integers is divisible by 9.
12 8 Chapter 1 28 Problem If x 1,n N prove that equals 1 1+x x x n x8 1+x 2n 1 x 1 + 2n+1 1 x 2n Problem Is it true that for every natural number n the quantity n 2 + n+41 is a prime? Prove or disprove! 30 Problem Give an example of an assertion which is not true for any positive integer, yet for which the induction step holds. 31 Problem Give an example of an assertion which is true for the first two million positive integers but fails for every integer greater than Problem Prove by induction on n that a set having n elements has exactly 2 n subsets. 33 Problem Prove that if n is a natural number, is always an integer. n 5 /5+n 4 /2+n 3 /3 n/30 34 Problem (Halmos) ) Every man in a village knows instantly when another s wife is unfaithful, but never when his own is. Each man is completely intelligent and knows that every other man is. The law of the village demands that when a man can PROVE that his wife has been unfaithful, he must shoot her before sundown the same day. Every man is completely law-abiding. One day the mayor announces that there is at least one unfaithful wife in the village. The mayor always tells the truth, and every man believes him. If in fact there are exactly forty unfaithful wives in the village (but that fact is not known to the men,) what will happen after the mayor s announcement? 35 Problem 1. Let a 1,a 2,...a n be positive real numbers with a 1 a 2 a n = 1. Use induction to prove that a 1 + a 2 + +a n n, with equality if and only if a 1 = a 2 = = a n = Use the preceding part to give another proof of the Arithmetic-Mean-Geometric-Mean Inequality. 3. Prove that if n > 1, then 4. Prove that if n > 1 then (2n 1) < n n. n (n+1) 1/n 1 < n. 5. Prove that if n > 1 then n < n 1 1 (n+1) 1/n + 1 n Given that u, v, w are positive, 0 < a 1, and that u+v+w = 1, prove that 1 u a 1 v a 1 w a 27 27a+9a 2 a Let y 1,y 2,...,y n be positive real numbers. Prove the Harmonic-Mean- Geometric-Mean Inequality: n n y1 y 2 y n. y 1 y 2 y n 8. Let a 1,...,a n be positive real numbers, all different. Set s = a 1 + a 2 + +a n. (a) Prove that (n 1) s < 1 1. s a 1 r n r a 1 r n r (b) Deduce that 4n s < 1 a r (s a r ) n 1 < n 1. a r 1 r n 1 r n 36 Problem Suppose that x 1,x 2,...,x n are nonnegative real numbers with x 1 + x 2 + +x n 1/2. Prove that (1 x 1 )(1 x 2 ) (1 x n ) 1/2. 37 Problem Given a positive integer n prove that there is a polynomial T n such that cosnx = T n (cosx) for all real numbers x. T n is called the n-th Tchebychev Polynomial. 38 Problem Prove that for all natural numbers n > 1. 1 n n n > 13 24
13 Fibonacci Numbers 9 39 Problem In how many regions will a sphere be divided by n planes passing through its centre if no three planes pass through one and the same diameter? 40 Problem (IMO 1977) Let f, f : N N be a function satisfying f(n+1) > f(f(n)) for each positive integer n. Prove that f(n) = n for each n. 41 Problem Let F 0 (x) = x,f(x) = 4x(1 x),f n+1 (x) = F(F n (x)),n = 0,1,... Prove that 1 (Hint: Let x = sin 2 θ.) 0 F n (x)dx = 22n 1 2 2n Fibonacci Numbers The Fibonacci numbers f n are given by the recurrence f 0 = 0, f 1 = 1, f n+1 = f n 1 + f n, n 1. (1.5) Thus the first few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21,.... A number of interesting algebraic identities can be proved using the above recursion. 42 Example Prove that Solution: We have Summing both columns, as desired. 43 Example Prove that f 1 + f f n = f n+2 1. f 1 = f 3 f 2 f 2 = f 4 f 3 f 3 = f 5 f 4. f n = f n+2 f n+1. f 1 + f f n = f n+2 f 2 = f n+2 1, f 1 + f 3 + f f 2n 1 = f 2n. Solution: Observe that Adding columnwise we obtain the desired identity. f 1 = f 2 f 0 f 3 = f 4 f 2 f 5 = f 6 f 4... f 2n 1 = f 2n f 2n 2 44 Example Prove that Solution: We have Thus f f f 2 n = f n f n+1. f n 1 f n+1 = ( f n+1 f n )( f n + f n 1 ) = f n+1 f n f 2 n + f n+1 f n 1 f n f n 1. f n+1 f n f n f n 1 = f 2 n,
14 10 Chapter 1 which yields f f f 2 n = f n f n Theorem (Cassini s Identity) f n 1 f n+1 f 2 n = ( 1) n, n 1. Proof: Observe that f n 1 f n+1 fn 2 = ( f n f n 2 )( f n + f n 1 ) fn 2 = f n 2 f n f n 1 ( f n 2 f n ) = ( f n 2 f n fn 1) 2 Thus if v n = f n 1 f n+1 f 2 n, we have v n = v n 1. This yields v n = ( 1) n 1 v 1 which is to say 46 Example (IMO 1981) Determine the maximum value of f n 1 f n+1 f 2 n = ( 1) n 1 ( f 0 f 2 f 2 1 ) = ( 1) n. m 2 + n 2, where m, n are positive integers satisfying m, n {1, 2, 3,..., 1981} and (n 2 mn m 2 ) 2 = 1. Solution: Call a pair (n,m) admissible if m,n {1,2,...,1981} and (n 2 mn m 2 ) 2 = 1. If m = 1, then (1,1) and (2,1) are the only admissible pairs. Suppose now that the pair (n 1,n 2 ) is admissible, with n 2 > 1. As n 1 (n 1 n 2 ) = n 2 2 ± 1 > 0, we must have n 1 > n 2. Let now n 3 = n 1 n 2. Then 1 = (n 2 1 n 1 n 2 n 2 2) 2 = (n 2 2 n 2 n 3 n 2 3) 2, making (n 2,n 3 ) also admissible. If n 3 > 1, in the same way we conclude that n 2 > n 3 and we can let n 4 = n 2 n 3 making (n 3,n 4 ) an admissible pair. We have a sequence of positive integers n 1 > n 2 >..., which must necessarily terminate. This terminates when n k = 1 for some k. Since (n k 1,1) is admissible, we must have n k 1 = 2. The sequence goes thus 1,2,3,5,8,...,987,1597, i.e., a truncated Fibonacci sequence. The largest admissible pair is thus (1597, 987) and so the maximum sought is Let τ = be the Golden Ratio. Observe that τ 1 =. The number τ is a root of the quadratic equation 2 2 x 2 = x+1. We now obtain a closed formula for f n. We need the following lemma. 47 Lemma If x 2 = x+1,n 2 then we have x n = f n x+ f n 1. Proof: We prove this by induction on n. For n = 2 the assertion is a triviality. Assume that n > 2 and that x n 1 = f n 1 x+ f n 2. Then x n = x n 1 x = ( f n 1 x+ f n 2 )x = f n 1 (x+1)+ f n 2 x = ( f n 1 + f n 2 )x+ f n 1 = f n x+ f n 1 48 Theorem (Binet s Formula) The n-th Fibonacci number is given by n = 0,2,... f n = n n
15 Practice 11 Proof: The roots of the equation x 2 = x+1 are τ = τ n = τ f n + f n 1 and 1 τ = 1 5. In virtue of the above lemma, 2 and Subtracting from where Binet s Formula follows. (1 τ) n = (1 τ) f n + f n 1. τ n (1 τ) n = 5 f n, 49 Example (Cesàro) Prove that Solution: Using Binet s Formula, n k=0 n k 2 k f k = = = n k=0 n k 2 k f k = f 3n. n k=0 n k 2 k τk (1 τ) k 5 n k=0 n 1 5 n k=0 n k τ k 1 ((1+2τ) n (1+2(1 τ)) n ). 5 k 2 k (1 τ) k As τ 2 = τ + 1,1+2τ = τ 3. Similarly 1+2(1 τ) = (1 τ) 3. Thus as wanted. The following theorem will be used later. n k=0 n k 2 k f k = 1 5 (τ) 3n +(1 τ) 3n = f 3n, 50 Theorem If s 1, t 0 are integers then f s+t = f s 1 f t + f s f t+1. Proof: We keep t fixed and prove this by using strong induction on s. For s = 1 we are asking whether f t+1 = f 0 f t + f 1 f t+1, which is trivially true. Assume that s > 1 and that f s k+t = f s k 1 f t + f s k f t+1 for all k satisfying 1 k s 1. We have f s+t = f s+t 1 + f s+t 2 by the Fibonacci recursion, = f s 1+t + f s 2+t trivially, = f s 2 f t + f s 1 f t+1 + f s 3 f t + f s 2 f t+1 by the inductive assumption = f t ( f s 2 + f s 3 )+ f t+1 ( f s 1 + f s 2 ) rearranging, = f t f s 1 + f t+1 f s by the Fibonacci recursion. This finishes the proof. Practice
16 12 Chapter 1 51 Problem Prove that 52 Problem Prove that 53 Problem Prove that f n+1 f n f n 1 f n 2 = f 2n 1, n > 2. f 2 n+1 = 4 f n f n 1 + f 2 n 2, n > 1. f 1 f 2 + f 2 f f 2n 1 f 2n = f 2 2n. 54 Problem Let N be a natural number. Prove that the largest n such that f n N is given by log N n = log Problem Prove that f 2 n + f 2 n 1 = f 2n Problem Prove that 62 Problem Prove that 63 Problem Prove that 64 Problem Prove that 1 arctan = π/4. f n=1 2n+1 f n lim n τ n = 1. 5 f n+r lim = τ r. n f n 1 f n k=0 2 k = 2+ f 2 n 2. f 2 n 56 Problem Prove that if n > 1, f 2 n f n+l f n l = ( 1) n+l f 2 l. Deduce that k=0 1 f 2 k = Problem Prove that n k=1 58 Problem Prove that f 2k = n=2 (n k) f 2k+1. n k=0 1 f n 1 f n+1 = Problem (Cesàro) Prove that 66 Problem Prove that n k=0 n k f k = f 2n. Hint: What is 1 f n 1 f n 1 f n f n+1? is a rational number. n=1 f n 10 n 59 Problem Prove that 60 Problem Prove that n=1 f n f n+1 f n+2 = 1. 1/ f 2 n = 4 τ. n=0 67 Problem Find the exact value of 1994 k=1( 1) k 1995 k. k f 68 Problem Prove the converse of Cassini s Identity: If k and m are integers such that m 2 km k 2 = 1, then there is an integer n such that k = ± f n,m = ± f n+1.
17 Pigeonhole Principle Pigeonhole Principle The Pigeonhole Principle states that if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons. This apparently trivial principle is very powerful. Let us see some examples. 69 Example (Putnam 1978) Let A be any set of twenty integers chosen from the arithmetic progression 1,4,...,100. Prove that there must be two distinct integers in A whose sum is 104. Solution: We partition the thirty four elements of this progression into nineteen groups {1},{52}, {4,100}, {7,97}, {10,94},... {49,55}. Since we are choosing twenty integers and we have nineteen sets, by the Pigeonhole Principle there must be two integers that belong to one of the pairs, which add to Example Show that amongst any seven distinct positive integers not exceeding 126, one can find two of them, say a and b, which satisfy b < a 2b. Solution: Split the numbers {1, 2, 3,..., 126} into the six sets {1,2},{3,4,5,6},{7,8,...,13,14},{15,16,...,29,30}, {31,32,...,61,62} and {63,64,...,126}. By the Pigeonhole Principle, two of the seven numbers must lie in one of the six sets, and obviously, any such two will satisfy the stated inequality. 71 Example Given any set of ten natural numbers between 1 and 99 inclusive, prove that there are two disjoint nonempty subsets of the set with equal sums of their elements. Solution: There are = 1023 non-empty subsets that one can form with a given 10-element set. To each of these subsets we associate the sum of its elements. The maximum value that any such sum can achieve is = 945 < Therefore, there must be at least two different subsets that have the same sum. 72 Example No matter which fifty five integers may be selected from prove that one must select some two that differ by 10. {1,2,...,100}, Solution: First observe that if we choose n+1 integers from any string of 2n consecutive integers, there will always be some two that differ by n. This is because we can pair the 2n consecutive integers into the n pairs {a+1,a+2,a+3,...,a+2n} {a+1,a+n+1},{a+2,a+n+2},...,{a+n,a+2n}, and if n+1 integers are chosen from this, there must be two that belong to the same group. So now group the one hundred integers as follows: and {1,2,...20},{21,22,...,40}, {41,42,...,60}, {61,62,...,80} {81,82,...,100}. If we select fifty five integers, we must perforce choose eleven from some group. From that group, by the above observation (let n = 10), there must be two that differ by 10.
18 14 Chapter 1 73 Example (AHSME 1994) Label one disc 1, two discs 2, three discs 3,..., fifty discs 50. Put these = 1275 labeled discs in a box. Discs are then drawn from the box at random without replacement. What is the minimum number of discs that must me drawn in order to guarantee drawing at least ten discs with the same label? Solution: If we draw all the = 45 labelled 1,..., 9 and any nine from each of the discs 10,..., 50, we have drawn = 414 discs. The 415-th disc drawn will assure at least ten discs from a label. 74 Example (IMO 1964) Seventeen people correspond by mail with one another each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there at least three people who write to each other about the same topic. Solution: Choose a particular person of the group, say Charlie. He corresponds with sixteen others. By the Pigeonhole Principle, Charlie must write to at least six of the people of one topic, say topic I. If any pair of these six people corresponds on topic I, then Charlie and this pair do the trick, and we are done. Otherwise, these six correspond amongst themselves only on topics II or III. Choose a particular person from this group of six, say Eric. By the Pigeonhole Principle, there must be three of the five remaining that correspond with Eric in one of the topics, say topic II. If amongst these three there is a pair that corresponds with each other on topic II, then Eric and this pair correspond on topic II, and we are done. Otherwise, these three people only correspond with one another on topic III, and we are done again. 75 Example Given any seven distinct real numbers x 1,...x 7, prove that we can always find two, say a,b with 0 < a b 1+ab < 1 3. Solution: Put x k = tana k for a k satisfying π 2 < a k < π 2. Divide the interval ( π 2, π ) into six non-overlapping subintervals of 2 equal length. By the Pigeonhole Principle, two of seven points will lie on the same interval, say a i < a j. Then 0 < a j a i < π 6. Since the tangent increases in ( π/2,π/2), we obtain as desired. 0 < tan(a j a i ) = tana j tana i 1+tana j tana i < tan π 6 = 1 3, 76 Example (Canadian Math Olympiad 1981) Let a 1,a 2,...,a 7 be nonnegative real numbers with If a 1 + a a 7 = 1. M = max 1 k 5 a k + a k+1 + a k+2, determine the minimum possible value that M can take as the a k vary. Solution: Since a 1 a 1 + a 2 a 1 + a 2 + a 3 and a 7 a 6 + a 7 a 5 + a 6 + a 7 we see that M also equals max {a 1,a 7,a 1 + a 2,a 6 + a 7,a k + a k+1 + a k+2 }. 1 k 5 We are thus taking the maximum over nine quantities that sum 3(a 1 + a 2 + +a 7 ) = 3. These nine quantities then average 3/9 = 1/3. By the Pigeonhole Principle, one of these is 1/3, i.e. M 1/3. If a 1 = a 1 + a 2 = a 1 + a 2 + a 3 = a 2 + a 3 + a 4 = a 3 +a 4 +a 5 = a 4 +a 5 +a 6 = a 5 +a 6 +a 7 = a 7 = 1/3, we obtain the 7-tuple (a 1,a 2,a 3,a 4,a 5,a 6,a 7 ) = (1/3,0,0,1/3,0,0,1/3), which shows that M = 1/3. Practice
19 Practice Problem (AHSME 1991) A circular table has exactly sixty chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest possible value of N? Answer: Problem Show that if any five points are all in, or on, a square of side 1, then some pair of them will be at most at distance 2/2. 79 Problem (Eötvös, 1947) Prove that amongst six people in a room there are at least three who know one another, or at least three who do not know one another. 80 Problem Show that in any sum of non-negative real numbers there is always one number which is at least the average of the numbers and that there is always one member that it is at most the average of the numbers. 81 Problem We call a set sum free if no two elements of the set add up to a third element of the set. What is the maximum size of a sum free subset of {1,2,...,2n 1}. Hint: Observe that the set {n+1,n+2,...,2n 1} of n+1 elements is sum free. Show that any subset with n+2 elements is not sum free. 82 Problem (MMPC 1992) Suppose that the letters of the English alphabet are listed in an arbitrary order. 1. Prove that there must be four consecutive consonants. 2. Give a list to show that there need not be five consecutive consonants. 3. Suppose that all the letters are arranged in a circle. Prove that there must be five consecutive consonants. 83 Problem (Stanford 1953) Bob has ten pockets and forty four silver dollars. He wants to put his dollars into his pockets so distributed that each pocket contains a different number of dollars. 1. Can he do so? 2. Generalise the problem, considering p pockets and n dollars. The problem is most interesting when Why? n = (p 1)(p 2) Problem No matter which fifty five integers may be selected from {1,2,...,100}, prove that you must select some two that differ by 9, some two that differ by 10, some two that differ by 12, and some two that differ by 13, but that you need not have any two that differ by Problem Let mn + 1 different real numbers be given. Prove that there is either an increasing sequence with at least n+1 members, or a decreasing sequence with at least m+1 members. 86 Problem If the points of the plane are coloured with three colours, show that there will always exist two points of the same colour which are one unit apart. 87 Problem Show that if the points of the plane are coloured with two colours, there will always exist an equilateral triangle with all its vertices of the same colour. There is, however, a colouring of the points of the plane with two colours for which no equilateral triangle of side 1 has all its vertices of the same colour. 88 Problem Let r 1,r 2,...,r n,n > 1 be real numbers of absolute value not exceeding 1 and whose sum is 0. Show that there is a non-empty proper subset whose sum is not more than 2/n in size. Give an example in which any subsum has absolute 1 value at least n Problem Let r 1,r 2,...,r n be real numbers in the interval [0,1]. Show that there are numbers ε k,1 k n,ε k = 1,0,1 not all zero, such ε k r k that n k=1 n 2 n. 90 Problem (USAMO, 1979) Nine mathematicians meet at an international conference and discover that amongst any three of them, at least two speak a common language. If each of the mathematicians can speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language. 91 Problem (USAMO, 1982) In a party with 1982 persons, amongst any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?
20 16 Chapter 1 92 Problem (USAMO, 1985) There are n people at a party. Prove that there are two people such that, of the remaining n 2 people, there are at least n/2 1 of them, each of whom knows both or else knows neither of the two. Assume that knowing is a symmetrical relationship. 93 Problem (USAMO, 1986) During a certain lecture, each of five mathematicians fell asleep exactly twice. For each pair of these mathematicians, there was some moment when both were sleeping simultaneously. Prove that, at some moment, some three were sleeping simultaneously. 94 Problem Let P n be a set of en! +1 points on the plane. Any two distinct points of P n are joined by a straight line segment which is then coloured in one of n given colours. Show that at least one monochromatic triangle is formed. (Hint: e = 1/n!.) n=0
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