Let $a, b$, and $c$ be integers. Suppose that $a$ and $b$ are coprime and that $a^2 +b^2 = c^3$. Show that $a\not\equiv b \pmod 2$ and that $c$ is odd

Question

What I'm thinking is this: Let $a = 2n+b$, then the equation becomes $2(2n^2+2bn+b^2)=c^3$. I'm stuck here.

After this, how can I show that $a+bi$ and $a-bi$ are coprime Gaussian integers?

https://proofwiki.org/wiki/Parity_of_Smaller_Elements_of_Primitive_Pythagorean_Triple — lab bhattacharjee, May 11 '17 at 18:04
https://math.stackexchange.com/questions/1714742/three-questions-about-the-form-x2-pm-3y2-z3-and-a-related-lemma/1715457#1715457 https://math.stackexchange.com/questions/1561774/diophantine-equation-x2-y2-z3/1562011#1562011 — individ, May 11 '17 at 18:19

score 3 · Answer 1 · answered May 11 '17 at 18:07

3

As $(a,b)=1$ both can not be even together

If both are odd $(2A+1)^2+(2B+1)^2\equiv2\pmod8$

$\implies c^3$ is even $\implies c$ is even $\implies c^3\equiv0\pmod8\not\equiv2$

So, $a,b$ must be of opposite parity.

WLOG $a$ is even and $b$ is odd

$\implies a^2$ is even, $b^2$ is odd

answered May 11 '17 at 18:07

lab bhattacharjee

1

Ah, so the sum of the squares of two odd numbers can never be the cube of an integer? – turkeyhundt May 11 '17 at 18:12

1 Answers1