# 26 is unique between a square (5^2) and a cube (3^3)

by al-mahed
Tags: cube, square, unique
 P: 258 i.e, $$a^{2} \pm 2 = b^{3}$$ has (25,27) as the only solution. Now, can you prove it?
 PF Gold P: 1,059 I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.
P: 258
 Quote by robert Ihnot I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.
Hi Robert, I'm sorry I didn't clarify that a, b must be positive integers.

 P: 361 26 is unique between a square (5^2) and a cube (3^3) Are you asking for help in proving the statement or challenging us to give a proof?
P: 258
 Quote by Petek Are you asking for help in proving the statement or challenging us to give a proof?
second option :)
 P: 39 It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD). For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z. So a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and 1 = d*(3c^2 - 2d^2). So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2. Same situation for a^2 - 2 = b^3.
PF Gold
P: 1,059
 Quote by hochs It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD). For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z. So a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and 1 = d*(3c^2 - 2d^2). So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2. Same situation for a^2 - 2 = b^3.
Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
P: 39
 Quote by robert Ihnot Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
Yea, I don't know why the post in that link is using both a + sqrt(-2) = w^3 and a - sqrt(-2) = t^3 (conjugate). You only need the first, you get just as much information just more quickly.
 P: 361 To go from $$(a + \sqrt{-2})(a - \sqrt{-2}) = b^3$$ to $$a + \sqrt{-2} = (c + d\sqrt{-2})^3$$ don't we need that $a + \sqrt{-2}$ and $a - \sqrt{-2}$ are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD($a + \sqrt{-2}, a - \sqrt{-2}$) divides $2\sqrt{-2}$. I think that's why he has to consider several cases.
P: 39
 Quote by Petek To go from $$(a + \sqrt{-2})(a - \sqrt{-2}) = b^3$$ to $$a + \sqrt{-2} = (c + d\sqrt{-2})^3$$ don't we need that $a + \sqrt{-2}$ and $a - \sqrt{-2}$ are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD($a + \sqrt{-2}, a - \sqrt{-2}$) divides $2\sqrt{-2}$. I think that's why he has to consider several cases.
Yes, that's right. So what I wrote above is not entirely correct, but one gets the idea - just factor and use UFD-ness of Z[sqrt(-2)].

More importantly, does anyone know how to use tex here without having to type itex and \itex everytime I want it? for example i prefer to , just much easier.
 P: 361 The pdf file linked at the beginning of this thread strongly suggests that itex, /itex and its variants are the only format that works here.
 P: 258 Hi folks, I followed a different direction. first notice that a and b must have the same parity, so first case: 1) a, b = even $$(2k)^{2}\pm 2 = (2n)^{3}$$ $$4k^{2}\pm 2 = 8n^{3}$$ so, $$2k^{2}\pm 1= 4n^{3}$$ absurd, since even number cannot be equal to odd number, so second case: 2) a, b = odd $$(2k+1)^{2}\pm 2 = (2n+1)^{3}$$ 2.1) + 2 $$4k(k+1)+3 = 8n^{3}+12n^{2}+6n+1$$ wich means $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4$$ $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k$$ and $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (K+1)$$ so $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4$$ $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod k$$ and $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod (K+1)$$ given $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .... + a_{i}x^{m_{i}}$$ and $$g(x) = b_{0} + b_{1}x + b_{2}x^{2} + b_{3}x^{3} + ....+ b_{j}x^{n_{j}}$$ we know that $$f(x)\equiv\ g(x)\mod W$$ $$\Leftrightarrow$$ $$a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W$$ lets make $$f(x) = 4n^{3}+6n^{2}+3n$$ and $$g(x) = 1$$ finally (skip obvious manipulations) $$12\equiv\ 0 \mod 4$$ $$12\equiv\ 0 \mod k$$ and $$12\equiv\ 0 \mod (K+1)$$ k must be = 2 2.2) - 2 (...) (after the same steps) finally $$14\equiv\ 0 \mod 4$$ what is untrue Any remarks would be apreciated.
 P: 258 Is there anything not clear, or perhaps wrong? I apreciate some feedback
 P: 258 Thinking again, I suspect I misunderstood the theorem found here: http://planetmath.org/encyclopedia/P...ongruence.html and jumped to a conclusion prematurely. $$f(x)\equiv\ g(x)\mod W$$ $$a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W$$ Seems that the conclusion above is wrong. Anyway, I'd preciate if anyone could make the theorem more clear to me, and make any comments about what I've posted :) thank you!
 P: 258 $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4$$ $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k$$ and $$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (k+1)$$ so $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4$$ $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod k$$ and $$4n^{3}+6n^{2}+3n\equiv\ 1 \mod (k+1)$$ that's also wrong, but do not detroy the following argument because $$24\equiv\ 0 \mod 4$$ $$24\equiv\ 0 \mod k$$ and $$24\equiv\ 0 \mod (k+1)$$ still would imply that k must be = 2

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