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1 PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the infinite descent. Fermat, Introduction The method of descent is a technique developed by Fermat for proving certain equations have no (or few) integral solutions. The idea is to show that if there is an integral solution to an equation then there is another integral solution that is smaller in some way. Repeating this process and comparing the sizes of the successive solutions leads to an infinitely decreasing sequence a 1 > a 2 > a 3 > of positive integers, and that is impossible. Let s take a look at two examples. Example 1.1 (Euler). We will show the equation x 3 + 2y 3 + 4z 3 = 0 has no solution in integers other than the obvious solution (0, 0, 0). Assume there is a solution (x, y, z) (0, 0, 0), so at least one of x, y, and z is not 0. The equation tells us x 3 is even, so x is even. Write x = 2x. Then 8x 3 + 2y 3 + 4z 3 = 0. Dividing by 2 and rearranging terms, we get y 3 + 2z 3 + 4x 3 = 0. This is just like our original equation, with (x, y, z) replaced by (y, z, x ). Since y is now playing the role previously played by x, the argument used before on x shows y is even. Writing y = 2y, substituting this in, and removing a common factor of 2, we get z 3 + 2x 3 + 4y 3 = 0. Therefore z is even, so z = 2z. Substituting this in and simplifying, x 3 + 2y 3 + 4z 3 = 0. Thus (x, y, z ) fits the original equation and at least one of x, y or z is nonzero (corresponding to whichever of x, y, and z is nonzero). Since 0 < max( x, y, z ) = (1/2) max( x, y, z ), we have produced a smaller integral solution measured by the maximum absolute value, which is a positive integer. This process can be repeated infinitely often, leading to a contradiction. The same proof shows for any prime p that the equation x 3 + py 3 + p 2 z 3 = 0 has no integral solution other than (0, 0, 0). Indeed, if (x, y, z) fits the equation then p x 3, so p x and we can proceed exactly as in the special case p = 2. In Section 2 we will give proofs by descent that certain numbers are irrational. In Section 3 we will show the equation a 4 + b 4 = c 4 (a special case of Fermat s Last Theorem) has no solution in positive integers using descent. In Section 4 we will use descent to show certain equations have no solution in nonconstant rational functions. In a positive direction, descent will be used in Section 5 to show any prime p such that 1 mod p is a sum of two squares. In Section 6 we will argue by descent that for any integer k > 0 other than 1 or 3, the equation x 2 + y 2 + z 2 = kxyz has no integral solutions (x, y, z) besides (0, 0, 0). While descent may appear to be something like reverse induction, it is not as widely applicable in the whole of mathematics as induction. Descent is nevertheless quite central to some important developments in number theory. 1
2 2 KEITH CONRAD 2. Irrationality by descent Here is the usual proof that 2 is irrational, expressed using the idea of descent. Example 2.1. We assume 2 is rational, so 2 = a/b with positive integers a and b. Squaring both sides and clearing the denominator, 2b 2 = a 2. (This is an equation we want to show is not solvable in positive integers.) Since 2 a 2, 2 a. Write a = 2a for some positive integer a, so 2b 2 = 4a 2, which is the same as b 2 = 2a 2. Thus 2 b 2, so 2 b. Write b = 2b, so 4b 2 = 2a 2, which is the same as 2b 2 = a 2. Since a and b are positive, we have 2 = a /b, so a 2 = b = a b. Since b = 2b and both b and b are positive, 0 < b < b, so we started with one rational expression for 2 and found another rational expression with a smaller (positive) denominator. Now we can repeat this process and obtain a sequence of rational expressions for 2 with decreasing positive denominators. This can t go on forever, so we have a contradiction. The way this proof usually is written starts with 2 = a/b where the fraction is in lowest terms. Then the fact that a = 2a and b = 2b, as shown in the theorem, is a contradiction since it means the fraction wasn t in lowest terms. The method of descent bypassed having to put the fraction in lowest terms, obtaining a contradiction in a different way. Let s take a look at another proof by descent that 2 is irrational. We assume 2 is rational. Since 1 < 2 < 2, we can write m (2.1) 2 = 1 + n, where m and n are positive integers with 0 < m/n < 1, so 0 < m < n. Squaring both sides of (2.1) and clearing the denominator, 2n 2 = n 2 + 2mn + m 2, so m 2 = n 2 2mn = n(n 2m). Since m 2 and n are positive, so is n 2m, and m n = n 2m m. This lies between 0 and 1, by the definition of m/n, so 0 < n 2m < m. We have reached the descent step: the fractional part m/n of 2 has been written as a fraction (n 2m)/m with a smaller denominator than before: 0 < m < n. We can repeat this process again and again, eventually reaching a contradiction. This proof by descent that 2 is irrational is not the same as the proof by descent in Example 2.1, since it does not use anything about even and odd numbers. It also generalizes nicely to other square roots. Theorem 2.2. If d Z + and d is not a perfect square then d is irrational. Proof. (Dedekind, 1858) Suppose d is rational. Since d is not a perfect square, its square root lies between two consecutive integers. Let l be the integer such that l < d < l + 1. (Note l is uniquely determined by d.) Write d = l + m n,
3 PROOFS BY DESCENT 3 where m and n are positive integers with 0 < m/n < 1, so 0 < m < n. Squaring both sides and clearing the denominator, dn 2 = n 2 l 2 + 2mnl + m 2, so m 2 = nq, where q = n(d l 2 ) 2ml. Since m 2 and n are positive, q is positive. Then m/n = q/m, so m d = l + n = l + q m. Since q/m = m/n, 0 < q/m < 1, so 0 < q < m. The fraction q/m has a smaller (positive) denominator than m/n, so from one representation d l = m/n we get another representation d l = q/m with a smaller (positive) denominator. This leads to a contradiction by repeating this process enough times. Here is another proof of Theorem 2.2, using descent in Z 2 rather than in Z. The argument is taken from [3]. Proof. Set A = ( 0 1 d 0 ). Its characteristic polynomial is det(λi 2 A) = λ 2 d, with an eigenvalue d and associated eigenvector ( ) d 1. Assuming d is rational, write d = a/b with nonzero integers a and b. Any scalar multiple of an eigenvector is an eigenvector, and ( ) ( d 1 = a/b ) ( 1 can be scaled to a ) ( b. This is also an eigenvector of A: A a ( b) = db ) ( a = d a b). Let l be the integer such that l < d < l + 1. Then ( ) a (A li 2 ) = ( ) ( ) a a d l = ( ( ) a d l), b b b b where d l lies between 0 and 1. The integral vector ( a b) is an eigenvector of the integral matrix A li 2 with eigenvalue between 0 and 1. Since ( a b) is an eigenvector of A li2, it is also an eigenvector of (A li 2 ) r for any r 1, with eigenvalue ( d l) r : ( ) a (A li 2 ) r = ( ( ) a d l) r. b b On the left side, for any r 1 we have a vector in Z 2 since A has integer entries and a, b, and l are integers. On the right side we have a nonzero vector (since a, b, and d l are nonzero) and it is getting arbitrarily small as r grows since d l < 1. So we have a sequence of nonzero vectors in Z 2 with length shrinking to 0 (the descent idea). This is impossible, so we have a contradiction. We can extend the same proof to cube roots, using descent in Z 3. Theorem 2.3. If d Z and d is not a perfect cube then 3 d is irrational. Proof. Suppose 3 d = a/b with nonzero integers a and b. Let A = 0 0 d 1 0 0, v = a2 ab, b 2 so det(λi 3 A) = λ 3 d and Av = (a/b)v = 3 dv. Let l Z satisfy l < 3 d < l + 1, so (A li 3 )v = ( 3 d l)v. Then (2.2) (A li 3 ) r v = ( 3 d l) r v
4 4 KEITH CONRAD for all r 1. Since v Z 3 and A has integer entries, the left side of (2.2) is a vector in Z 3. Since v 0 and 0 < 3 d l < 1, the right side of (2.2) is nonzero and its length is tending to 0 as r grows. Thus, as r, the nonzero vectors (A li 3 ) r v are a sequence in Z 3 with length shrinking to 0. This is impossible, so 3 d must be irrational. Remark 2.4. In a similar way one can deal with higher roots: if d Z and k 2 (with d > 0 if k is even) and d is not a kth power in Z then k d is irrational. Just assume k d = a/b is rational and use the k k matrix and vector a k 1 a k 2 b A = ( 0 d I k 1 0 ), v =. b k Fermat s Last Theorem for n = 4 We will use descent to prove the exponent 4 case of Fermat s Last Theorem: the equation a 4 + b 4 = c 4 has no solution in positive integers. Fermat proved something more general, allowing a square and not just a fourth power on the right side. Theorem 3.1 (Fermat). There is no solution to the equation x 4 + y 4 = z 2 in positive integers. In particular, the equation a 4 + b 4 = c 4 has no solution in positive integers. Proof. We will use the parametrization of primitive Pythagorean triples, so let s recall that: a primitive solution to a 2 + b 2 = c 2 where a, b, and c are positive integers with b even is a = k 2 l 2, b = 2kl, c = k 2 + l 2, where k > l, (k, l) = 1, and k l mod 2. Assume there is a solution to x 4 + y 4 = z 2 where x, y, and z are positive integers. If p is a common prime factor of x and y then p 4 z 2, so p 2 z. Then we can cancel the common factor of p 4 throughout and get a similar equation with smaller positive values of x, y, and z. Doing this enough times, we may suppose that (x, y) = 1. Then (x, z) = 1 and (y, z) = 1 too. We will find a second positive integer solution (x, y, z ) with (x, y ) = 1 that is smaller in a suitable sense. Since x 4 + y 4 = z 2 and (x, y) = 1, at least one of x and y is odd. They can t both be odd, since otherwise z 2 2 mod 4, which has no solution. Without loss of generality, say x is odd and y is even. Then z is odd. Since (x 2 ) 2 + (y 2 ) 2 = z 2, (x 2, y 2, z) is a primitive Pythagorean triple with y 2 the even term, so by the formula for primitive triples we can write (3.1) x 2 = k 2 l 2, y 2 = 2kl, z = k 2 + l 2, where k > l > 0 and (k, l) = 1 (also k l mod 2, but we don t need this). The first equation in (3.1) says x 2 + l 2 = k 2. Since (k, l) = 1, (x, l, k) is another primitive Pythagorean triple. Since x is odd, using the formula for primitive Pythagorean triples once again tells us (3.2) x = a 2 b 2, l = 2ab, k = a 2 + b 2, where a > b > 0 and (a, b) = 1. The second equation in (3.1) now says y 2 = 4(a 2 + b 2 )ab.
5 PROOFS BY DESCENT 5 Since y is even, ) 2 = (a 2 + b 2 )ab. ( y 2 Since (a, b) = 1, the three factors on the right are pairwise relatively prime. They are all positive, so their product being a square means each one is a square: (3.3) a = x 2, b = y 2, a 2 + b 2 = z 2, where x, y, and z can all be taken as positive. From (a, b) = 1, (x, y ) = 1. The last equation in (3.3) can be rewrittten as x 4 + y 4 = z 2, so we have another solution to our original equation with (x, y ) = 1. Now we compare z to z. Since 0 < z z 2 = a 2 + b 2 = k k 2 < z, measuring the size of positive integer solutions (x, y, z) by the size of z leads to a contradiction by descent. Remark 3.2. At the end of the proof a simple estimate showed z > z 2. We can also get a formula for z in terms of x, y, and z that explains this inequality. By (3.1), (3.2), and (3.3), z = k 2 + l 2 = (a 2 + b 2 ) 2 + (2ab) 2 = z 4 + 4x 4 y 4, so in fact z > z 4, not just z > z 2 as we found before. Let s write x and y in terms of x, y, and z too. From (3.2) and (3.3), x = a 2 b 2 = x 4 y 4 and y 2 = 2kl = 2(a 2 + b 2 )(2ab) = 4z 2 (x y ) 2, so y = 2x y z. This formula for y shows x, y, and z are all less than y, so 0 < max(x, y, z ) < y max(x, y, z). Using max(x, y, z) rather than z to measure the size of a solution (x, y, z) is another way to get a contradiction for Theorem 3.1 by descent. Our proof of Theorem 3.1 used the parametric formula for primitive Pythagorean triples twice. For a proof that does not explicitly use this parametrization, see [1, pp ]. If we apply the descent technique for x 4 + y 4 = z 2 to a 4 + b 4 = c 4, with a fourth power on the right side, then the proof breaks down. The reason is that the descent step will not return another solution of a 4 + b 4 = c 4 ; the smaller c that comes out will only show up as a square, not a 4th power. So the extra generality of dealing with x 4 + y 4 = z 2 is essential for the descent to work as above. Elementary number theory books that discuss Fermat s Last Theorem for exponent 4 introduce the equation x 4 + y 4 = z 2 out of the blue, like we did, as if it were the most natural thing in the world to look at this equation instead of a 4 +b 4 = c 4. Of course it isn t. Fermat was actually thinking about x 4 + y 4 = z 2 not in order to solve a 4 + b 4 = c 4 but for an entirely different reason, and it was natural to consider the equation for that other problem. See Appendix A for more details. Now we present a number of corollaries to Theorem 3.1, concerning solvability of certain equations in integers or rationals. None of the proofs (which are mostly short) will involve descent. They are presented here simply to show Theorem 3.1 has uses other than Fermat s Last Theorem for exponent 4. Corollary 3.3. Any rational solution to x 4 + y 4 = z 2 has x or y equal to 0.
6 6 KEITH CONRAD Proof. Assume x and y are both nonzero. Then z 2 > 0, so z 0 too. Write x = a/d, y = b/d, and z = c/d with a, b, c, d Z and d > 0. Then a, b, and c are nonzero. Clearing the denominator in x 4 + y 4 = z 2, we have a 4 + b 4 = (cd) 2. Changing signs if necessary, a, b, and cd are positive. Then we have a contradiction with Theorem 3.1. Corollary 3.4. The only rational solutions to y 2 = x are (0, ±1). Proof. Use Corollary 3.3 to see x = 0. Corollary 3.5. The only rational solutions to 2y 2 = x 4 1 are (±1, 0). Proof. Squaring both sides, 4y 4 = x 8 2x Add 4x 4 to both sides and divide by 4 to get y 4 + x 4 = ((x 4 + 1)/2) 2. Since x 0 in the original equation, we can divide by x 4 to get (y/x) = ((x 4 + 1)/2x 2 ) 2. By Corollary 3.4, y/x = 0, so y = 0 and therefore x = ±1. Corollary 3.6. The integral solutions of x 4 y 4 = 2z 2 are (x, ±x, 0). Proof. If y = 0 then x = z = 0 since 2 is irrational. If y 0, then divide by y 4 to get (x/y) 4 1 = 2(z/y 2 ) 2. By Corollary 3.5, z/y 2 = 0, so z = 0 and therefore y = ±x. Corollary 3.7. The only rational solutions to y 2 = x 3 4x are (0, 0), (±2, 0). Proof. There is a bijection between solutions of y 2 = x 3 4x with x 0 and solutions to v 2 = u by ( ) ( ) y (x, y) 2x, y2 + 8x 2 4x 2, (u, v) v u 2, 4u v u 2. Since any rational solution to v 2 = u has u = 0, any rational solution to y 2 = x 3 4x has y = 0, so x = 0 or x = ±2. Corollary 3.8. The only rational solution to y 2 = x 3 + x is (0, 0). Proof. Writing the equation as y 2 = x(x 2 + 1), we see x = 0 if and only if y = 0. Assume there is a rational solution other than (0, 0) so x 0 and y 0. From the equation, x must be positive. Write x and y in reduced form as x = a/b and y = c/d where b and d are positive. Clearing denominators in (c/d) 2 = (a/b) 3 + a/b, we get b 3 c 2 = d 2 (a 3 + ab 2 ). Therefore d 2 b 3 c 2. Since (c, d) = 1, d 2 b 3. Also b 3 d 2 (a 3 + ab 2 ). Since (a, b) = 1, b 3 is relatively prime to a 3 + ab 2, so b 3 d 2. Thus b 3 = d 2, so by unique factorization b = t 2 and d = t 3 for some positive integer t. Then (a, t) = 1 and (c, t) = 1. In the equation y 2 = x 3 + x with x = a/t 2 and y = c/t 3, we get c 2 = a 3 + t 4 a = a(a 2 + t 4 ) after clearing the denominator. Since (a, t) = 1, a and a 2 + t 4 are relatively prime and positive. Their product is a square, so each factor is a square: a = u 2, a 2 + t 4 = v 2. Thus u 4 + t 4 = v 2. By Theorem 3.1, u or t is 0. Since t 0, u = 0 so x = 0 and then y = 0. Remark 3.9. Conversely, Corollary 3.8 implies Theorem 3.1. If x 4 + y 4 = z 2 in positive integers then multiplying through by x 2 /y 6 gives us (x/y) 6 + (x/y) 2 = (xz/y 3 ) 2, so Y 2 = X 3 + X for X = (x/y) 2 and Y = xz/y 3. Since X is a nonzero rational number, we have a contradiction with Corollary 3.8.
7 PROOFS BY DESCENT 7 Here is another theorem about fourth powers and squares proved by Fermat using descent. Theorem 3.10 (Fermat). There is no solution to x 4 y 4 = z 2 in positive integers. Proof. We will argue by descent in a similar style to the proof of Theorem 3.1. In particular, we will use the formula for primitive Pythagorean triples twice. Since now we have z 2 +y 4 = x 4 while in Theorem 3.1 we had x 4 +y 4 = z 2, the roles of x 2 and z basically get interchanged. For example, we will use descent on x 2 (or equivalently, on x) rather than on z as we did in Theorem 3.1. Assume x 4 y 4 = z 2 with x, y, and z in Z +. There must be a solution with x, y, and z pairwise relatively prime (see the start of the proof of Theorem 3.1; the same argument there applies here), so we suppose this is the case. Since x 4 y 4 > 0, x > y. There are two cases to consider: z odd and z even. Case 1: z is odd. Since z 2 + y 4 = x 4 and z is odd, y must be even. (Otherwise z 2 + y mod 4, but 2 is not a 4th power modulo 4.) Since (x, y) = 1, (z, y 2, x 2 ) is a primitive Pythagorean triple with y 2 the even term, so the formula for primitive Pythagorean triples says (3.4) z = k 2 l 2, y 2 = 2kl, x 2 = k 2 + l 2, where k > l > 0, (k, l) = 1, and k l mod 2. The third equation in (3.4) says (k, l, x) is a Pythagorean triple. Since (k, l) = 1, this triple is primitive. One of k or l is odd and the other is even. If k is odd, the formula for primitive Pythagorean triples says (3.5) k = a 2 b 2, l = 2ab, x = a 2 + b 2, where a > b > 0, (a, b) = 1, and a b mod 2. If l is odd the formula says (3.6) l = a 2 b 2, k = 2ab, x = a 2 + b 2, where a > b > 0, (a, b) = 1, and a b mod 2. Using whichever of (3.5) or (3.6) is correct (depending on the parity of k and l), the second equation in (3.4) becomes (3.7) y 2 = 4(a 2 b 2 )ab. Since y is even, we can divide by 4 (in Z): ( y ) 2 = (a 2 b 2 )ab. 2 Since (a, b) = 1, the three factors on the right are pairwise relatively prime. They are all positive, so their product being a square means each one is a square: (3.8) a = x 2, b = y 2, a 2 b 2 = z 2, where x, y, and z can all be taken as positive. From (a, b) = 1, (x, y ) = 1. The last equation in (3.8) can be rewrittten as x 4 y 4 = z 2, so we have another solution to our original equation. Moreover, z 2 = a 2 b 2 is odd, so our new solution again has an odd square on the right and we are still in Case 1. Now we compare x to x: 0 < x x 2 = a < a 2 + b 2 = x. Since x < x, by descent we have a contradiction. Case 2: z is even. (This has no analogue in the proof of Theorem 3.1.) Since y 4 + z 2 = x 4, we have a primitive Pythagorean triple (y 2, z, x 2 ) with even z. Thus y 2 = m 2 n 2, z = 2mn, x 2 = m 2 + n 2,
8 8 KEITH CONRAD where m and n are positive and (m, n) = 1. Multiplying the first and third equations, (xy) 2 = m 4 n 4, with xy odd. This expresses a square as the difference of two fourth powers, with the square being odd, so by Case 1 we have a contradiction. Remark In Case 1 we can solve for x, y, and z in terms of x, y, and z. From (3.5) or (3.6), x = a 2 + b 2. This becomes, by (3.8), x = x 4 + y 4. From (3.7) and (3.8), y 2 = 4(a 2 b 2 )ab = 4z 2 (x 2 y 2 ) = (2x y z ) 2, so Lastly, by (3.4), (3.5) or (3.6), and (3.8), y = 2x y z. z = k 2 l 2 = ±((a 2 b 2 ) 2 (2ab) 2 ) = ±(z 4 4x 4 y 4 ), so z = z 4 4x 4 y 4. From the formula y = 2x y z we get 0 < max(x, y, z ) < y max(x, y, z), so using max(x, y, z) rather than x as a measure of the size of a positive integer solution is another way of reaching a contradiction in Case 1 by descent. This parallels Remark 3.2. Theorems 3.1 and 3.10 together lead to the following two results. Corollary There is no Pythagorean triple in which two of the terms are squares. Proof. Such a triple would give a solution in positive integers to either x 4 + y 4 = z 2 (the two legs are squares) or x 4 = y 4 + z 2 (a leg and hypotenuse are squares), but such solutions do not exist by Theorems 3.1 and Many primitive Pythagorean triples have just one term a square. See Table 1. a b c Table 1. Pythagorean triples with a square term Corollary The only m Z + such that m is a fourth power is 1. Proof. Since m = m(m + 1)/2, we are trying to solve m(m + 1)/2 = n 4. Clearing the denominator, m(m + 1) = 2n 4. Since m and m + 1 are relatively prime, {m, m + 1} = {x 4, 2y 4 } for some positive integers x and y, which must be relatively prime. Therefore x 4 2y 4 = ±1. Rewrite as y 4 = (x 4 1)/2 and square both sides followed by a little algebra to get ( x (3.9) y 8 ± x 4 4 ) 2 ± 1 =. 2
9 PROOFS BY DESCENT 9 The right side is an integer since x is odd from x 4 = 2y 4 ± 1. Equation (3.9) expresses a sum or difference of positive fourth powers as a perfect square. Using Theorems 3.1 and 3.10, we see this is impossible for a sum, and for a difference the square must be 0, so x = 1 (since x > 0). Therefore m = 1. Now we present consequences of Theorem 3.10 alone, paralleling the consequences of Theorem 3.1. Corollary Any rational solution to x 4 y 4 = z 2 has y or z equal to 0. Proof. Similar to Corollary 3.3. Corollary The only rational solutions to y 2 = x 4 1 are (±1, 0). Proof. Similar to Corollary 3.4. Corollary The only rational solutions to 2y 2 = x are (±1, ±1). Proof. Squaring both sides, 4y 4 = x 8 + 2x Subtract 4x 4 from both sides and divide by 4 to get y 4 x 4 = ((x 4 1)/2) 2. Since x 0 in the original equation, we can divide by x 4 to get (y/x) 4 1 = ((x 4 1)/2x 2 ) 2. By Corollary 3.15, (x 4 1)/2x 2 = 0, so x = ±1 and therefore y = ±1. Corollary The integral solutions of x 4 + y 4 = 2z 2 are (x, ±x, ±x 2 ). Proof. If y = 0 then x = z = 0 since 2 is irrational. If y 0, then divide by y 4 to get (x/y) = 2(z/y 2 ) 2. By Corollary 3.16, x/y = ±1 and z/y 2 = ±1, so y = ±x and z = ±y 2 = ±x 2. Corollary The only rational solutions to y 2 = x 3 + 4x are (0, 0) and (2, ±4). Proof. A bijection between solutions of y 2 = x 3 +4x with x 0 and solutions of v 2 = u 4 1 is given by ( ) ( ) y (x, y) 2x, y2 8x 2 4x 2, (u, v) u 2 v, 4u u 2. v Any rational solution to v 2 = u 4 1 has v = 0, so any rational solution to y 2 = x 3 + 4x has x = 0 or y 2 = 8x. The second case implies x 3 = 4x, so x = ±2. Only x = 2 leads to a solution, (x, y) = (2, ±4). Corollary The only rational solutions to y 2 = x 3 x are (0, 0) and (±1, 0). Proof. This stands in relation to x 4 y 4 = z 2 in the same way that y 2 = x 3 + x does to x 4 + y 4 = z 2 in Corollary 3.8. Details are left to the reader as an exercise. Remark In the other direction, Corollary 3.19 implies Theorem The argument is like that in Remark Equation with no rational function solution There are many rational solutions to x 2 +y 2 = 1, like (x, y) = (3/5, 4/5) or (5/13, 12/13). These solutions are nearly all described by a single parametric formula: x = 2t/(1 + t 2 ) and y = (1 t 2 )/(1 + t 2 ) for some t Q. Try t = 1/3, 1/4, and 1/5. Algebraically, we have an identity of rational functions ( ) 2t 2 ( ) 1 t t t 2 = 1
10 10 KEITH CONRAD and specializing t to rational numbers gives rational solutions to x 2 + y 2 = 1. 1 However, it is not true that rational solutions to other equations can always be fit into a parametric formula using rational functions. For example, the equation y 2 = x 3 2 has infinitely many rational solutions (two of them are (x, y) = (3, 5) and (129/100, 383/1000)), but there are no rational functions f(t) and g(t) in Q(t) such that g(t) 2 = f(t) 3 2 other than constant functions, which parametrize only one solution (so really do not provide a parametric formula at all). To prove negative results like this we will use descent on the degree in a hypothetical nonconstant solution. Several times we will need the following lemma. Lemma 4.1. If f(t) and g(t) in C[t] are relatively prime and fg = in C[t] then f = and g = in C[t]. Proof. By unique factoriatization, the multiplicity of any irreducible factor of f g is even. Since f and g are relatively prime, it follows that the multiplicity of any irreducible factor of f is even and likewise for g. Therefore f and g are squares up to a nonzero scaling factor. Every nonzero complex number is the square of a complex number, so f and g are squares. This lemma easily extends by induction on the number of terms to a product of any finite number of polynomials in C[t] that are pairwise relatively prime. Theorem 4.2. For distinct complex numbers r, r, r, any solution to the equation y 2 = (x r)(x r )(x r ) in rational functions x = f(t) and y = g(t) in C(t) is a constant solution: f(t) and g(t) are in C. The example y 2 = x 3 2 that we discussed above is the special case where r, r, and r are the three cube roots of 2. Proof. Our argument is adapted from [2, pp ]. Assume there is a solution in rational functions: x = p 1 (t)/p 2 (t) and y = q 1 (t)/q 2 (t) where p 1 (t), p 2 (t), q 1 (t), and q 2 (t) are polynomials in C[t]. Without loss of generality we can assume (p 1, p 2 ) = 1 and (q 1, q 2 ) = 1 in C[t]. Substituting the formulas for x and y into the equation y 2 = (x r)(x r )(x r ), we have q1 2 q2 2 = ( ) ( ) ( ) p1 p1 r r p1 r p 2 p 2 p 2 so after clearing denominators = (p 1 rp 2 )(p 1 r p 2 )(p 1 r p 2 ) p 3, 2 (4.1) p 3 2q 2 1 = (p 1 rp 2 )(p 1 r p 2 )(p 1 r p 2 )q 2 2. Since (p 1, p 2 ) = 1, the factors p 1 rp 2, p 1 r p 2, and p 1 r p 2 are relatively prime to p 2 (why?). Thus by (4.1), p 3 2 q2 2. Since (q 1, q 2 ) = 1, q2 2 p3 2, so p3 2 = cq2 2 for some nonzero c C. Since c is a square in C, p 3 2 is a square in C[t]. That implies p 2 is a square by unique factorization in C[t]. Write p 2 = f2 2. Substituting cq2 2 for p3 2 in (4.1) and cancelling q2 2 from both sides, cq 2 1 = (p 1 rp 2 )(p 1 r p 2 )(p 1 r p 2 ). 1 This method produces all the rational solutions to x 2 + y 2 = 1 except for (0, 1), but we will not show that here.
11 PROOFS BY DESCENT 11 The factors on the right side are pairwise relatively prime since (p 1, p 2 ) = 1 (why?) and the numbers r, r, and r are distinct, so by an extension of Lemma 4.1 to a product of three terms, p 1 rp 2, p 1 r p 2, and p 1 r p 2 are all squares in C[t]. Since p 2 = f 2 2, p 1 rf 2 2 =, p 1 r f 2 2 =, p 1 r f 2 2 =. Writing the first equation as p 1 rf2 2 = f 1 2, the second and third equations become (4.2) f 2 1 (r r)f 2 2 =, f 2 1 (r r)f 2 2 =, where r r and r r are nonzero and distinct. We want to show f 1 and f 2 are constant. Now we set up our descent statement, based on (4.2): we will show for all distinct a and b in C that relatively prime polynomials g 1 and g 2 in C[t] that satisfy (4.3) g 2 1 ag 2 2 =, g 2 1 bg 2 2 = must both be constant. Note (4.2) is a special case of this. Assume for some a and b that there is a solution (g 1, g 2 ) to (4.3) where g 1 or g 2 is not constant. In (4.3), write a = c 2 and b = d 2 with c and d in C. Since a b, c ±d. We can rewrite (4.3) as (4.4) (g 1 + cg 2 )(g 1 cg 2 ) =, (g 1 + dg 2 )(g 1 dg 2 ) =. A common factor of g 1 + cg 2 and g 1 cg 2 is a factor of both g 1 and g 2 (why?), so it is constant since (g 1, g 2 ) = 1. Therefore the factors on the left side of the first equation in (4.4) are relatively prime, and the product of the factors is a square, so by Lemma 4.1 (4.5) g 1 + cg 2 = h 2 1, g 1 cg 2 = h 2 2, where h 1 and h 2 are relatively prime. Adding and subtracting the equations in (4.5), g 1 = (h h2 2 )/2 and g 2 = (h 2 1 h2 2 )/(2c). Since g 1 or g 2 is not constant, h 1 or h 2 is not constant. Arguing in a similar way with the second equation in (4.4), (4.6) g 1 + dg 2 =, g 1 dg 2 =. Substituting the formulas for g 1 and g 2 in terms of h 1 and h 2 into (4.6), ( d ) h ( 1 d ) ( h =, 1 d ) h ( 1 + d ) h 2 2 =. 2 c 2 c 2 c 2 c The numbers 1 ± d/c are nonzero since c ±d, so we can divide by the coefficient of h 2 1 : h d/c 1 + d/c h2 2 =, h d/c 1 d/c h2 2 =. Set a = (1 d/c)/(1 + d/c) and b = (1 + d/c)/(1 d/c) = 1/a. Both a and b are nonzero, a b, and (4.7) h 2 1 a h 2 2 =, h 2 1 b h 2 2 =. From (4.5), 2 deg h 1 max(deg g 1, deg g 2 ) and 2 deg h 2 max(deg g 1, deg g 2 ). Therefore (4.8) 0 < max(deg h 1, deg h 2 ) 1 2 max(deg g 1, deg g 2 ). We can now repeat the argument leading from (4.3) to (4.7) with h 1, h 2, a, b in place of g 1, g 2, a, b. Each repetition leads to a new version of (4.8) where the maximum degree in the new solution is positive and smaller than the maximum degree in the previous solution. By descent this leads to a contradiction (positive degrees can t strictly drop forever), so any
12 12 KEITH CONRAD solution to (4.3) in relatively prime polynomials g 1 and g 2 must be a constant solution. What this tells us in (4.2) is that f 1 and f 2 are both constant, so p 1 = f1 2 + rf 2 2 and p 2 = f2 2 are constant. That makes x = p 1 /p 2 constant, so the equation y 2 = (x r)(x r )(x r ) implies that y 2 = (q 1 /q 2 ) 2 is constant. Therefore the rational function y = q 1 /q 2 is constant. Corollary 4.3. Let F (x) be a polynomial with coefficients in C of degree 3 or 4 that has distinct roots. If x and y in C(t) satisfy y 2 = F (x) then x C and y C. Proof. First suppose F (x) has degree 3. In factored form F (x) = c(x r)(x r )(x r ) where the roots r, r, and r are distinct and c 0. 2 In C, c is a square, so the equation (4.9) y 2 = c(x r)(x r )(x r ) can be scaled to Y 2 = (x r)(x r )(x r ) where Y = y/ c. If there is a solution to (4.9) where x, y C(t) then Y C(t). By Theorem 4.2, x C and Y C, so y C. Next suppose F (x) has degree 4 with distinct roots r, r, r, and r. In factored form where c 0. Suppose we can solve F (x) = c(x r)(x r )(x r )(x r ), (4.10) y 2 = c(x r)(x r )(x r )(x r ) where x, y C(t). The roots of F are distinct, so without loss of generality x r. Dividing through (4.10) by (x r ) 4, y 2 ( ) ( ) ( ) x r x r x r (x r ) 4 = c x r x r x r ( ) ( ) ( ) = c 1 + r r x r 1 + r r x r 1 + r r x r. Set X = 1/(x r ) and Y = y/(x r ) 2, so X and Y are in C(t). Then Y 2 = c(1 + (r r)x)(1 + (r r )X)(1 + (r r )X), where the right side is cubic in X with distinct roots. By the cubic case we already discussed, X C and Y C, so x C and y C. 5. Sums of two squares We have used descent to prove negative theorems, which say certain equations have no (or very few) solutions of a certain type. Fermat s initial applications of descent had this flavor. 3 He later found descent could be used to prove positive theorems like this. Theorem 5.1. For a prime p, if 1 mod p then we can write p as a sum of two squares: p = x 2 + y 2 for some x and y in Z. Proof. Write 1 s 2 mod p. Since s 2 ( s) 2 mod p we can take 1 s p/2, so s = pd with d Z. Since s p 2 /4 + 1 < p 2, we have 0 < d < p. So a multiple of p is a sum of two squares and the multiplying factor d is less than p. If d = 1 we are done. 2 The difference between this case and Theorem 4.2 is that we now allow a leading coefficient besides 1. 3 Almost none of Fermat s proofs are known in detail, but he did include in letters that statements of some propositions that he established with descent.
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BAART Berlin
BAART Berlin provides comprehensive opioid addiction medication maintenance therapy supported by individual and group counseling, addressing recovery from many different angles, including relapse prevention, wellness, women's issues and peer support groups. We provide thorough medication monitoring, as our doctor is in the office five days a week. We work closely with our community partners to provide wrap around services and treat the individual as a whole. Our goal is to provide quality care that is unique to each patient, and we emphasize that success looks different for each person.
BAART Berlin coordinates with local physicians who prescribe Suboxone to facilitate smooth transitions between levels of care, and provide support for patients who need to stabilize in our program in order to return to their physicians care again.
- Methadone & Buprenorphine Maintenance
- Individualized Substance Use Disorder Counseling
- Group Therapy
- Addiction Education
- Relapse Prevention
- Case Management
- Coordinated Pregnancy Treatment Resources
"The whole BAART Berlin program is great. The staff are great people that work well together to help their patients."
–BAART Berlin Patient
Phone: 802.223.2003
Fax: 802.223.2235
M – F: 6:00am – 2:00pm
S & S: 6:00am – 10:00am
M – F: 6:00am – 2:00pm
M – F: 5:30am – 9:45am &
11:00am – 1:00pm
S & S: 6:00am – 10:00am
Holidays: 6:00am – 10:00am
- Private Pay: Cash, Money Order , Debit Card
- Medicaid
- BlueCross/BlueShield
- MVP (Vermont Private Insurance)
- Sliding Fee Scale
- Medicaid Transportation(GMTA/RCT depending on residence)
- Hospital Hill Route Bus
© BAART Programs 2016

subset: commoncrawl


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On DVD
Into the Blue
- No Rating Available
Movie Review
Six-hundred billion dollars. That's the worth of the sunken treasure some experts estimate still lies on the floors of the world's oceans. And a passion-rich but cash-poor young American named Jared is determined to discover his share in the clear, shallow—and shark infested—waters off the Bahamas. Jared's live-in girlfriend, Sam, indulges his unlikely dream of striking it rich, but mostly she just loves him and being with him.
The couple's idyllic, if simple, life together is interrupted by the arrival of Jared's brother, Bryce, a shady but successful defense attorney, and his latest girlfriend (of 14 hours), Amanda. While free diving, the four of them make two momentous discoveries: the wreckage of a mythic pirate ship and the fuselage of a plane that's obviously crashed recently—and is full of cocaine.
To claim the ship they've discovered, which is potentially worth hundreds of millions of dollars, Jared and Co. have to confidently identify it and file a claim. Bryce and Amanda, however, are more interested in a sure thing: the millions of dollars of watertight cocaine bricks resting nearby. When the less scrupulous couple tries to sell some purloined coke to a local dealer named Reyes, they are sucked into a swirling vortex of underworld—and underwater—violence they never anticipated ... drawing Jared and Sam in as well.
Positive Elements
Jared and Amada both steadfastly resist Bryce and Amanda's insistence that they go after the stash of cocaine they've discovered. The group needs money to finance their salvage operation on the wreckage they've discovered, and that need generates strong temptation. But both Jared and Sam refuse to give into it.
Their moral character contrasts with that of Bryce and Amanda. Bryce rationalizes, "Winners make the rules, and losers live by them." For him, the cocaine represents a means to an end, and he's completely unconcerned with any effects selling the drug might have on its end users. Sam, on the other hand, values relationship above money. She asks Jared, "Would you give up treasure for love? You should ask yourself what's missing in your life that money would get you." And when it appears to her that Jared has given in and "gone bad," she is unwilling to compromise herself even for his sake.
Several characters, including Jared, Sam and Bryce, put their lives on the line to save one another once things go awry. Though Bryce's poor decisions have put the group in jeopardy, he's willing to help his brother try to save Sam, who's held hostage by an island drug lord.
Spiritual Content
A man says that if he saw a shark, he would "run like Jesus of Nazareth on the water."
Sexual Content
The underwater director of photography for Into the Blue, Peter Zuccarini, said of the many submerged scenes where actors and actresses wear only skimpy swimsuits instead of wetsuits, "We felt it was important for the cast to stay streamlined so they could move well in the action sequences while remaining recognizable. And their sun-bronzed bodies looked fantastic against the intensely blue Bahamian water." So much for realism! In virtually every underwater scene, lingering, close-up camera shots focus on Sam's and Amanda's features—which are barely covered by teeny-weeny-usually-not-polka-dot bikinis—creating a sexually charged atmosphere during even mundane excavations. On land, Sam and Amanda sashay about wearing little more than they do in the ocean (going braless, wearing miniskirts and halter tops, etc.).
The guys get equal time onscreen, and the camera takes great care in repeatedly highlighting their chiseled abs, legs and pecs. In one scene part of Jared's backside is seen when his shorts ride low.
Amanda sunbathes topless (her back and the side of her breast are seen.) When she stands up, even more of her chest is exposed (but not from the front).
Wearing underwear, Sam climbs on top of Jared in bed and kisses him. (It's clear that the pair lives together and has a sexual relationship.) Elsewhere, Sam struggles with an assailant underwater who grabs one of her breasts and almost pulls her swimming suit off.
Sexually tinged banter includes, among other things, an oblique reference to oral sex. A scene in a nightclub shows both couples dancing and grinding suggestively against one another.
Violent Content
A man cuts his face on a Plexiglas panel during a plane crash; water violently cascades into the cockpit drowning the pilot, co-pilot and a drug runner. Later, the main characters discover their decomposing bodies. A cannon that the group tries to raise from the ocean floor falls, pinning Bryce's arm beneath it. Bryce gets into a fistfight with a drug dealer and his lackeys. Jared and Bryce fight a pair of thugs; Jared—who apparently has Aquaman-like abilities to stay underwater for long periods of time without oxygen—holds one of the men underwater with him, perhaps attempting to drown him.
But all that's just for starters. Once Jared and Bryce get tangled up with the local drug dealers, the movie amps up the violence. A drug lord hits Bryce in the face and puts a gun to Jared's head. Tiger sharks feast graphically on two unlucky characters, including one whose mauled leg is clearly shown. Not pretty.
A man is shot in the stomach and killed; another character is shot in the head. Dead bodies litter a drug dealer's yacht. Two people are shot with small harpoons—one in the face and another in the thigh. Sam breaks free from a ship's hold using an ax to hack off the hand of a dead man to whom she was handcuffed; we don't witness the blade's impact, but we briefly glimpse the corpse and its obviously detached hand. Sam slams a door on a thug's hand, mangling it badly; he smashes a window with his head and continues pursuing her. There's a lethal, massive, underwater explosion.
Crude or Profane Language
Characters utter the f-word twice and the s-word approximately 10 times. They take God's or Jesus' name in vain about a dozen times (including one use of "g--d--n"). "A--" is also used about 10 times, followed by a few instances each of "h---," "b--ch" and "p---ed." One character gives another the finger, and crude slang references to both male and female anatomy show up. When Sam discovers Amanda has stolen some cocaine, she derisively labels her a "coke whore."
Drug and Alcohol Content
Several scenes show Jared, Sam, Bryce, Amanda and others consuming mixed drinks and Bacardi Breezers. In one shot, they're downing mixed drinks and smoking cigars in a swimming pool. They also drink together at the dance club, where Amanda tries to buy cocaine from a dealer. When Bryce and Amanda try to sell some of the cocaine they've stolen, Bryce's frequent sniffles and exaggerated paranoid behavior imply that he's already sampled the goods. It's equally obvious that Amanda is an experienced cocaine user.
Other Negative Elements
Sam works as a shark wrangler at the Atlantis theme park (think Sea World). As she feeds nurse sharks, she talks to spectators breezily about how most species of sharks are not as dangerous as people believe. Later, divers frequently swim with sharks without much concern. The implication is that cool people swim with sharks.
Jared and Bryce repeatedly jump a speedboat's wake on jet skis, getting big air and even doing flips with their personal watercraft. They make these dangerous stunts look easy—easy enough that young, impressionable watchers might give it a try themselves ... and end up with a jet ski on top of them.
Deception comes almost as naturally to Bryce as breathing. He lies to Jared about his ability to finance their salvage operation. And his selfish character is revealed further when he quietly suggests to Jared that his new girlfriend, Amanda, shouldn't have any share in the treasure.
In a summer full of action movies long on special effects and short on story or character development, Into the Blue actually exceeded my admittedly modest expectations. The story takes several unexpected, dramatic turns, and its heroine, Sam, refuses to compromise her moral stance (at least when it comes to cocaine; sex seems perfectly OK with her).
Integrity isn't a given for Hollywood protagonists anymore. Unfortunately, neither are clothes. Suffice it to say this movie is a Maxim reader's dream come true. Though Jessica Alba has repeatedly told interviewers that she refuses to do nude scenes in movies such as this one and Sin City (in which she portrayed a stripper), she seems quite comfortable getting as close to nudity as possible, wearing only tiny, strategically placed triangles of cloth.
Some might call Alba's no-nudity/near-nudity stance inconsistent. She sees it as good business. She told Hollywood Life magazine, "I have to go to certain lengths to use sexuality to my advantage while guiding people to think the way I want them to. People think I'm an 'action babe' because Fox wanted to 'sell' [my television show] Dark Angel to a Maxim demographic, and they put me on the cover of that magazine. It wasn't that they put a gun to my head—it's just how they sell movies."
That attitude—and the body that goes with it—is abundantly on display in Into the Blue. The filmmakers' infatuation with Jessica's curves, combined with a boatload of profanity and grisly images, make plunging into the deep waters of Into the Blue a chilly proposition.
Pro-social Content
Objectionable Content
Summary Advisory
Plot Summary
Christian Beliefs
Other Belief Systems
Discussion Topics
Additional Comments/Notes
Episode Reviews
Readability Age Range
Paul Walker as Jared; Jessica Alba as Sam; Scott Caan as Bryce; Ashley Scott as Amanda; Josh Brolin as Bates; James Frain as Reyes; Dwayne Adway as Roy
John Stockwell ( Turistas, Blue Crush, Crazy/Beautiful)
Columbia Pictures


text:
- Which BAKFlip Tonneau is best for me? (PDF)
- Turn your truck into a workaholic workhorse with the BAK BAKFlip CS Tonneau Cover
- Includes both a BAKflip HD Tonneau and a BAK CS Rack for powerful cargo protection and easy oversized freight hauling
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- When the tailgate is locked, the BAKFlip CS tonneau cannot be opened
- The BAK Contractor Series Overhead Rack is built with stainless steel uprights and aluminum crossbars rated to hold a full 500 lbs
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Whether you're moving small boxes, hauling lengthy lumber or taking your kayak to the lake, there's always room for your freight when you equip your workhorse with the BAK BAKflip CS Tonneau Cover. Custom configured with both a BAKflip HD Tonneau and a matching BAK CS Overhead Rack, this two-in-one combo gives you double-decker versatility for work and play.
Security and utility are on full display with the BAKflip HD Tonneau Cover. This hard folding tonneau is built from aluminum panels with a rugged EPS core, making it strong enough to support up to 300 lbs of evenly distributed weight. Would-be bandits would need a blow torch to cut through, and smart seals help keep the elements from soaking your stuff. Plus, it folds all the way up against your back window when you need full bed access and bolts on easily with no drilling required.
For all the oversized stuff that won't fit down below, pack it on the BAK CS Overhead Rack. Integrated right into your BAKflip CS Truck Bed Cover, they move back and forth into position and lock down with a twist of the included T-Key. Sliding load locks make it easy to secure ladders, pipes and sports equipment up top. And the BAK Flip CS Rack is rated for a full 500 lbs. What's more, your BAK Contractor Series Cover and Rack are covered by a 2-Year Warranty.
- The tailgate lock on the truck is what secures the bed. If your truck does not have a tailgate lock, the cover may be opened by opening the tailgate
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12 Thrifty Trash to Treasure Crafts
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Latest Comments
"The Hushabye Shell Afghan is a beautiful blanket that would be a great shower gift. The pattern is an easy shell stitch and the colts are great for a boy or girl. I am sure that it will keep baby nice and warm. It is made with a nice four ply yarn and should work up fast. This one to save for a later date."
"The wavy rainbow baby blanket is one of the cutest that I have seen. The colors are so bright that a baby will enjoy looking at it. The blanket could also be worked up in other colors that would be great for that special little one. this is not for a beginner, so if this is your first try at crocheting you may not want to try it. Looking forward to working on this."
"The Springtime Lace Afghan Pattern is so cute. It will make a great baby blanket for that special baby. This could be made in many different colors that will really be pretty. The pattern reads really easy and a beginner could do this.I am looking forward to to this project in the very near future."
"he Cape Cod Afghan is really a cute pattern. It would be a great one for beginners. The pattern only calls for two skeins of yarn. The pattern reads really easy. This could also be used as a baby blanket and would look great in one of those baby colors. This is one pattern that I will definitely save for a later project down the road."
Water Bottle Room Divider
By: Heidi Borchers
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Add depth and dimension to any room with a Water Bottle Room Divider from Heidi Borchers. This craft is extremely inexpensive to make and looks store-bought. Recycled plastic bottle crafts are a great way to reduce, reuse and recycle!
- Plastic drink bottles
- Craft knife
- Scissors
- Acrylic paint (colors of your choice)
- Cosmetic sponge (for applying paint)
- Drill with 1/16″ drill bit
- Hole punch
- 22 gauge wire (colors of your choice)
- Round nose pliers
- Wire cutters
- Beads
- Mirror and tile pieces
- I love to find plastic drink bottles that have intriguing designs embossed onto the sides! I don't drink sodas so I ask my friends to keep their eyes open for fun bottles!
- Use a craft knife to puncture the side of the bottle and cut apart.
- Use scissors to cut out the side panel pieces and to clean up the cut edges of the bottle tops and the bottoms.
- Apply acrylic paint with cosmetic sponge by dabbing onto the inside (wrong side) of the plastic bottle. It may take several coats of paint to get a solid opaque coverage.
- Use the same painting procedure to paint the inside of the bottle bottom.
- Arrange the bottle pieces in desired pattern. Punch or drill holes into the pieces.
- To create the swirl in the end of the wire, use pliers to roll end and then continue to roll with fingers until you have created swirl of desired size. Cut wire to desired length to join two pieces together.
- Insert cut end of wire through punched hole in water bottle. Loop wire around swirl to secure.
- Add beads to wire.
- If desired add other embellishments to wire. I found these doves at the dollar store. I drilled through them to turn them into beads.
- Secure both ends of beaded and embellished wire to bottle pieces. Continue add more pieces to create desired length. I join my pieces together for wall hangings by looping the top wire onto a dowel stick.
- Here's a close up of a bottle cap. I glued mirror tiles and a flat backed marble to the top.
- On the bottle bottom, I glued a white flat backed marble.
- On the bottle side piece, I glued vintage look mirror tiles.
- You can arrange rows and rows of these to create wall art or keep going and make a room divider!
For a comprehensive video tutorial of this project, check out How to Make a Drink Bottle Room Divider.
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slaughterfamily200 2yahoocom
Dec 26, 2016
This is one of the neatest ideas for water bottles, the one in the instructions is a gatoraid bottle, which we aquire a lot of in a sports centered house hold. This could work great as a "window dressing over blinds in several different rooms woth color or other "themes" Love itttttt!
Toby K
Jul 26, 2016
This is an amazing idea! You'd never even know they were made from water bottles.
ladonasmile 955836 5
Apr 08, 2014
oh my gosh I love the idea, I always admire the design on those bottles and wonder what could be done with them cause it is nice, so what i see you did there to make a room divider is Genius. How bottles did you use to make one?
FaveCrafts Editor JL
Apr 08, 2014
Hi there, Since Heidi does not specify the amount of water bottles, I suggest contacting her via her designer profile. Thanks!
corinnerconstable 3586135
Jan 09, 2013
This is very clever! Great project for teens too! They can customize it for their own rooms. Thank you for sharing! Love being a Greenie crc 1/9/13
Jun 21, 2012
Ingenious! Kudos to Heidi. Couldn't be greener and so well done.
bibisaccessories 1 022820
May 06, 2012
luv it I'm always recycling these bottles thks for the inspiration
aprilowencreations 7012650
Feb 03, 2012
Great idea. Very inventive
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Artificial gill
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MakerBot Dissolvable Filament
List Price R 1,659
eB14 690
Discovery Miles 14 690
On Credit: R / month
Eligible for Cash on Delivery. Learn more
Free delivery available |
Free Exchanges & Returns for 30 days. Learn more
MakerBot Dissolvable Filament
The 1.75mm Dissolvable Filament from MakerBot is designed for use as a solid infill material with the MakerBot ABS Filament. The Dissolvable Filament can easily dissolve in a limonene bath in just 8 to 24 hours, depending on the size and geometry of the 3D print.
Limonene is a common, colorless, citrus- scented solvent available online and in grocery stores. The finished Dissolvable Filamaent print will carry a citrus scent for a few days.
The Dissolvable Filament adheres well to kapton tape on a heated build plate. Please make sure your build plate is level, clean, and oil- free in order to ensure high quality 3D prints.
The MakerBot Dissolvable Filament is comprehensively quality tested at both 100- micron and 200- micron layer resolutions on the MakerBot Replicator 2X to verify that the filament will perform every feature that the MakerBot MakerWare software offers and reliably slice and print every type of 3D printing available.
- Dissolvable
Compatible Printers:
- MakerBot Replicator 2X
- Glass Temperature : 212 Degrees F/100 Degrees C
- Melting Temperature : 320 Degrees F/160 Degrees C
- Nozzle Temperature : 446 Degrees F/230 Degrees C
- Spool Diameter: 8.00"/20.32 cm
- Spool Width: 3.00"/7.62 cm
- Spool Hub Hole: 2.00"/5.08 cm
- Diameter: 0.068"/0.175 cm
- 2.2 lbs/1 kg
Whatâ??s in the box:
- MakerBot Dissolvable Filament
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Are you looking for Commercial cleaners in QUEBEC? Hotfrog showcases more than 500 popular businesses related to Commercial cleaners in QUEBEC! Find a specific region or suburb using the left navigation.
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Spécialiste Intervention Sinistre Inc. is a group of skilled, experienced and passionate professionals committed to helping you recover from a disaste
Building Maintenance
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Here at Entretien ménager SMC, we take pride in providing the best services and satisfying our customer needs. We specialize in Cleaning Service, Cleaning Service, Cleaning Services, Cleaning, Window Cleaning, Home Clean…
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• Heavy duty cable w/ tough coating
• Resettable combination
• Quick release bracket included
• Length: 6 feet
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Central West 28-day Rainfall Forecast
- period
- 28-day
- 12-month
|Chance of rainfall within district
< 25%
25% to 50%
50% to 75%
≥ 75%
Issue Notes
The hemispheric long wave pattern has been relatively fast moving in recent weeks. There are six main troughs. Currently the most significant troughs are near the longitudes of South Africa, the southwest Indian Ocean, Western Australia, the south Pacific, the southeast Pacific, and the Atlantic Ocean.
Over southern and eastern Australia the cold front events with potential to bring widespread rain are now expected about 5 March to 9 March, 18 March to 22 March, and 22 March to 26 March. Rain events originating in the tropics and moving south are possible about 20 February to 24 February.
Over Western Australia the strongest cold fronts should occur about 6 March to 10 March, 18 March to 22 March, and 24 March to 28 March.
Forecast Explanation
This forecast is produced by a multi-model ensemble consisting of dynamical atmospheric models, which are forced by the latest observed atmosphere, ocean, land and ice conditions. The models are designed to simulate features of the real atmosphere, including the daily movement of long and short wave patterns in the Southern Hemisphere.
The future probability of rain in each district is estimated using output from the multi-model ensemble, combined with historical information about the difference between the model forecasts and observed rainfall.
In this deterministic framework the skill of the forecast tends to decrease with time, however the forecasts are updated daily to provide the latest estimates of rainfall probability out to 28 days.
Weather News
Ex-Tropical Cyclone Alfred: Borroloola bridge collapses, town split in half after NT flooding
The community of Borroloola has been physically split in half after heavy rainfall and flooding caused by ex-Tropical Cyclone Alfred partly collapsed a bridge.
Victoria told national energy operator 'under no circumstances' could it cut power during NSW heatwave
The Victorian Government has told the national energy operator "under no circumstances" could it cut power to regional centres in order to keep .
Ex-Tropical Cyclone Alfred makes landfall, weakens to tropical low
Tropical Cyclone Alfred has weakened to a tropical low as it nears landfall on the Gulf of Carpentaria's southern coast, but the Bureau of Meteorology warned the weather system remained powerful.


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John Biggs (yes, of TechCrunch) will choose five finalists to participate in the Make It in Brooklyn + Brooklyn Education Innovation Network pitch contest, where you can win $5,000 and other prizes such as One Month of Unlimited Commons Membership from WeWork (a $350 value) and a free Zipcar for Business membership with free driving credit (a $185 value)!
You get two minutes to pitch and four minutes to field questions from a panel of judges.
This contest is one in a series of pitch-offs we are holding over the next year in the run up to our 2nd annual Make It In Brooklyn Innovation Summit in late spring 2016.
Fill out the application below by Sunday, October 18th. Find out more info, visit makeitinbk.com. For more details about the Brooklyn Education Innovation Network, visit http://downtownbrooklyn.com/projects/college-town.
For questions about the pitch contest, send an email to email@example.com.


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iPhone apps
Lior Suchard: Supernatural Firestarter
Amaze your audience by performing this startling, interactive, mind-reading trick!Need an app that is sure to increase your popularity among friends, family, and even complete strangers? Hey, it's fun too! Supernatural Firestarter will grant you….. Read more
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5/5 stars
Lots of fun
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2/5 stars
Maybe if I could get it to work
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FREE FOR A LIMITED TIME ONLY! In conjunction with AppFreeway.com... The ultimate tool for getting ANY girl's phone number…Yep, ANY girl!Want to know the secret behind persuading any girl to give you… [more..]
Price: free |


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Five jobs in demand that don't require a four-year degree and promise bigger paychecks than some jobs requiring a college education.
Conventional wisdom has it that, if you want to get a good job that pays well, you need a college degree.
But if you're looking for jobs in demand, five of the professions expected to see the most new jobs this year not only don't require a four-year degree, they can pay more than the average $46,000 salary of a college graduate. Get one of these jobs, and you may be the envy of a humanities major with a job paying less than $30,000 a year or no job at all.
The U.S. Department of Labor lists the fastest growing professions and the professions expected to have the most openings by the amount of education required. Some of those jobs require only a high school diploma, some call for short-term vocational training and some require a two-year community college degree.
Health-related jobs are in high demand, and many medical jobs that don't require a four-year degree pay well. Moody's Economy recently reported that the number of jobs in education and health services will rise 2.2% this year, double the national average of 1.1 % for all jobs.
Great Jobs in Demand
Here's a look at the five jobs in demand that don't require a college degree, pay well, and are expected to have the most openings:
- Registered nurse. The U.S. Department of Labor projects 103,900 new openings this year. Although you can earn a bachelor's degree in nursing, community colleges offer two-year RN programs. Registered nurses make an average of $47,700 to $69,800 a year. Top earners, however, take home an average of more than $83,400 a year, according to the Nursing Schools website.
- Licensed practical nurse and licensed vocational nurse. This group will see 39,130 more positions this year. Training takes about a year. The median pay is about $39,000 a year with the top 10 percent of wage earners taking in more than $53,000, according to the U.S. Bureau of Labor Statistics.
- Computer support specialist. The feds estimate this high-paying field will grow 23,460 jobs in 2010. Computer specialists usually need an associate's degree from a community college. Their reward is a fat paycheck. The median annual salary for a network or computer systems administrator is $66,130, according to the U.S. Bureau of Labor Statistics. The top 10 percent break into six figures or about $104,000 a year.
- Hair dresser, hair stylist and cosmetologist. You can't outsource haircuts overseas, so hair dressers can expect more jobs to open up as the U.S. population grows. The job requires vocational training. Most make $30,000 to $50,000 a year but top earners can easily crack six figures. It helps to have people skills as many hairdressers rely on tips – which can be $300 or more a week.
- Auto service agent, technician and mechanic. This trade can be learned in high school or on the job. The average pay is about $40,000 a year, but many earn more.
Other jobs that pay well, don't require college degrees and are expected to have openings are insurance salesperson; heating, air conditioning and refrigeration mechanic and installer; real estate agent; welder, cutter, solderer and brazer; paralegal and legal assistant; dental hygienist; radiologic technologist and technician; and respiratory therapist.
Even for these jobs in demand, people may need to move to find work: Texas is strong in job growth while Michigan, no surprise, continues to shed jobs.


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23 Beaver Dam Drive,Sicklerville NJ, 08081
Sold Price
4 beds , 2 Full/1 Half baths
3,325 sq. ft
Status: Sold
Berkshire Hathaway HomeServices Fox & Roach, REALTORS
(866) 486-5636
Last Updated: 2/20/2017 12:34 PM
Description for 23 Beaver Dam Drive Sicklerville, NJ 08081
This property is being sold as is. The Buyers are responsible for any and all repairs, inspections, certifications and C/O. See attachment for PAS requirements and EFHM offer submittal information in MLS document section.
Listing Information
Wiltons Corner
Gas Water Heater, Public Water Service
Public Sewer
School Information
Room Information
Breakfast Area
First Floor Laundry
Breakfast Area
Interior Features
Central Air
Gas Heat
Full, Unfinished Basement
1, Living Room
Exterior / Lot Features
2 Garage Spaces, Attached Garage, Street Parking, Driveway
Community Pool
Sidewalks, Street Lights, Front Yard, Back Yard, Side Yard
Community Features
Community Pool
Financial Considerations
Inside AccessTM
Current market conditions for 23 Beaver Dam Drive
Active Market Trends
Source:MLS DATA
Median list price$154,997
Days on market112
Properties for sale280
Sold Market Trends
Source:MLS DATA
Median sold price$135,000
Sales to list price9,709.0%
Properties sold147
Median sq ft price$78
Arrows indicate 90 day trend
Listing Price History for 23 Beaver Dam Drive
Sicklerville Real Estate
Sicklerville is located in New Jersey. Sicklerville, New Jersey 08081 has a population of 50,544.
The median household income in Sicklerville, New Jersey 08081 is $77,889. The median household income for the surrounding county is $62,320 compared to the national median of $53,046. The median age of people living in Sicklerville 08081 is 35.6 years.
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