Solving $2x^4+2x^2y^2+y^4 = z^2$

Asked Dec 24 '16 at 01:29

Active Dec 24 '16 at 18:06

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Find all integer pairs $(x,y,z)$ that satisfy $$2x^4+2x^2y^2+y^4 = z^2.$$

We can rewrite the given equation as $(x^2+y^2)^2+(x^2)^2 = z^2$. Thus, $(x^2+y^2,x^2,z)$ must be a Pythagorean triple. How do we continue?

Also, for $x = 0$ we get $y^4 = z^2$ and so $z = \pm y^2$. Can we show that $x$ must be $0$?

edited Dec 24 '16 at 17:44

asked Dec 24 '16 at 01:29

user19405892

5

Pythagorean triples can be parametrized. – Mark Schultz-Wu Dec 24 '16 at 01:32
@user19405892 could you add the tag "diophantine-equations"? I think it's better when a question has more than one tag. – Xam Dec 24 '16 at 03:24
Doing modulo 3 and 5 should help in this question – Ariana Dec 24 '16 at 20:53
@ArianaGrande How?. It is easy to prove that the given equation has no solutions when $x$ is odd working modulo $4$ and $8,$ but how is it helpful to consider the eq. mod $3$ and/or $5$? – CIJ Dec 24 '16 at 21:31
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WTF?, why was this question marked as a duplicate of a question that was asked after this one? – CIJ Dec 25 '16 at 22:06

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