Solving $x^2+y^3=z^2$

Question

Solve $x^2+y^3=z^2$ in which x,y, and z are relatively prime and y is even.

I have done a similar proof for $x^2+y^2=z^2$; however, the $y^3$ significantly changes the method I used before.

My solution for that was

$x=v^2-u^2$, $y=2uv$, and $z=v^2+u^2$ where $u$ and $v$ are arbitrary relatively prime numbers in which one is even and one is odd.

See https://mathoverflow.net/questions/264648/primitive-integral-solutions-to-x2y3-z2 — Robert Z, May 18 '17 at 06:05

shrindle · Accepted Answer · 2017-05-19T03:44:07.240

0

I figured it out.

Similar to problem I did before,

$x=2u^3-v^3$ and $z=v^3+2u^3$ where $u$ and $v$ are arbitrary relatively prime numbers.

However y is the value that must change since it is a third power. Thus rather than taking a square root I must take a cube root.

Therefore, $y={2uv}$

The proof was very long and tedious. It took me many hours to get, but if you check that parametrization it does work.

edited May 19 '17 at 03:44

answered May 18 '17 at 06:30

shrindle

Isn't $y$ supposed to be an integer? – quasi May 18 '17 at 06:36
Does this actively solve as asked or just present a solution with out proof? How is the asker to understand if it is correct? Maybe it works for a few so? – marshal craft May 18 '17 at 06:38
Consider $x=y=3$, $z=6$. This parametrization does not cover all solutions. – Sophie May 22 '17 at 04:52

1 Answers1