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Mr.Math Group Title This group hasn't been active lately. So I'm going to post a set of fun problems bring it back to life. $\text{ Fun With Mr.Math #1}$ Find the integer solutions to the equation $9^x-3^x=y^4+2y^3+y^2+2y.$ 2 years ago 2 years ago Edit Question Delete Cancel Submit

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1. Tomas.A
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why wolframalpha doesn't give solution as x=0 and y=0? http://www.wolframalpha.com/input/?i=9%5Ex-3%5Ex%3Dy%5E4%2B2y%5E3%2By%5E2%2B2y+solve+integer

• 2 years ago
2. Mr.Math
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I don't know. I wish it doesn't give any solution at all :-D

• 2 years ago
3. sasogeek
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if x=0, y=an imaginary number

• 2 years ago
4. Tomas.A
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why?

• 2 years ago
5. Tomas.A
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9^0-3^0=1-1=0 and when y=0 then 0+0+0+0=0 same answer, no?

• 2 years ago
6. sasogeek
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factor out y, you'll end up with $$\ (y^2+2y)(y^2+1)=9^x-3^x$$ when x=0, $$\ y^2=-1, y=\sqrt{-1}$$

• 2 years ago
7. sasogeek
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but $$\ y^2+2y=0 => y=-2$$

• 2 years ago
8. Tomas.A
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y(y+2)(y^2+1)=0 so y=0 and y=-2 no?

• 2 years ago
9. sasogeek
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please write full statements lol, you're getting me lost by ... no? :P

• 2 years ago
10. Tomas.A
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you write $(y^2+2y)(y^2+1)=9^x-3^x$ if x=0 $(y^2+2y)(y^2+1)=0$ $y(y+2)(y^2+1)=0$ so $y=0\text{ and }y+2=0\text{ and }y^2+1=0$ so y=0 y=-2 and no real solutions

• 2 years ago
11. sasogeek
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=> my solutions is wrong? not sure... i guess i'll wait for to see the correct one :D

• 2 years ago
12. Tomas.A
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i don't care about solution because i dont have a clue how to solve but x=0 and y=0 is clearly true and u didn't prove me that it's wrong lol

• 2 years ago
13. sasogeek
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:P y=-2 and y=0 both work for x=0 and are true, but i wonder if they're the right solutions lol, Mr. Math, what are the right solutions?

• 2 years ago
14. Tomas.A
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so why wolframalpha does not show y=0?

• 2 years ago
15. sasogeek
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it isn't as smart as u ;)

• 2 years ago
16. asnaseer
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Here are my thoughts so far: First lets simplify the left hand side:$9^x-3^x=(3^2)^x-3^x=3^{2x}-3^x=3^x(3^x-1)$Then lets simplify the right hand side:$y^4+2y^3+y^2+2y=y(y+2)(y^2+1)$ We therefore have:$3^x(3^x-1)=y(y+2)(y^2+1)$Now the trivial solution found above is x=0 which leads to y=0 or y=-2 (since we are only interested in integer solutions). Another solution could also be $$x=-\infty$$ and y=0 or -2 but I assume we can rule out infinities. Now lets rearrange the equation to:$y^2+1=\frac{3^x(3^x-1)}{y(y+2)}$This means $$y(y+2)$$ must evenly divide into $$3^x(3^x-1)$$. Notice that $$3^x$$ is always an odd number and $$3^x-1$$ is always an even number. Case 1: If y is an even number, then y+2 will also be even. Therefore, if y is even, then $$y(y+2)$$ must evenly divide into $$3^x-1$$. So let $$y=2m$$ to get:$\frac{3^x-1}{4m(m+1)}\hspace{2cm}\text{must be an integer}$ Case 2: If y is an odd number, then y+2 will also be odd. Therefore, if y is odd, then $$y(y+2)$$ must evenly divide into $$3^x$$. So let $$y=2m+1$$ to get:$\frac{3^x}{(2m+1)(2m+3)}\hspace{2cm}\text{must be an integer}$

• 2 years ago
17. asnaseer
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Since 3 is a prime number, case 2 can be rejected because $$(2m+1)(2m+3)$$ can only evenly divide into $$3^x$$ if $$(2m+1)(2m+3)=3^k$$ which is impossible. So we are left with just Case 1 to solve.

• 2 years ago
18. asnaseer
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In case 1, if m>1, then $$4m(m+1)$$ will always contain a product of an even number and an odd number. However, we know the numerator $$3^x-1$$ is always even, therefore we can reject all values of m>1. So case 1 reduces to m=1 which gives y=2 (since $$y=2m$$ for this case). Substituting this back into the original equation gives:\begin{align} 3^x(3^x-1)&=y(y+2)(y^2+1)\\ &=2(2+2)(2^2+1)\\ &=2*4*5\\ &=40 \end{align}which does not give an integer solution for x. Therefore (I believe) the only solutions are: 1) x=0 and y=0 or -2 2) x=-$$\infty$$ and y=0 or -2

• 2 years ago
19. Mr.Math
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Great job everyone, some solution(s) is/are still missing though.

• 2 years ago
20. asnaseer
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hmmm - I re-checked my work and found some flaws in my logic. I have now proved that for y>0 y must be odd. And, by trial-and-error I have found one more solution: 3) x=1 and y=1 but I have not been able to prove that this is the only other solution :(

• 2 years ago
21. mukushla
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its immediate that $$x\ge0$$ letting $$x=0$$ gives 2 solution $(x,y)=(0,0)\ , \ (0,-2)$suppose that $$x>0$$ using asnaseer factoring$3^x(3^x-1)=y(y+2)(1+y^2)$RHS is divisible by $$3^x$$$y \equiv 0,1,2 \ \ \text{mod} \ 3 \\ y^2 \equiv 0,1 \ \ \text{mod} \ 3 \\ 1+y^2 \equiv 1,2 \ \ \text{mod} \ 3$so $$y(y+2)$$ is divisible by $$3^x$$ but we know that $$\gcd(y,y+2)=1,2$$ so exactly one of $$y$$ or $$y+2$$ is divisible by $$3^x$$. case 1 : $$y=3^xm$$ and $$m\neq0$$ and $$m$$ is not a multiple of 3$3^x-1=m(3^xm+2)(1+m^23^{2x})$RHS is greater than LHS in magnitude so no solution from here case 2 : $$y+2=3^xm$$ and $$m\neq0$$ and $$m$$ is not a multiple of 3$3^x-1=(3^xm-2)m(5+m^23^{2x}-4m3^x)$$3^x-1=m(3^xm-2)(5+(3^xm^2-4m)3^x)$again RHS is greater than LHS in magnitude unless $$x=1$$ so one solution from here$(x,y)=(1,1)$

• 2 years ago
22. mukushla
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actually for case 2 if $$x>1$$$|m(3^xm-2)(5+(3^xm^2-4m)3^x)|>|(5+(3^xm^2-4m)3^x)|>5+3^x$

• 2 years ago
23. mukushla
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only solutions$(x,y)=(0,0)\ ,\ (0,-2)\ , \ (1,1)$

• 2 years ago
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