Prove that the equation $x^4 = y^2 +z^2 +4$ has no integer solutions.

I believe I have proved it for the case when both $y$ and $z$ are of the same parity.

Case 1: When $y$ and $z$ are of the same parity. This proof is for the case when both are even. The other is similar. $$x^4 = y^2 + z^2 + 4 \Longleftrightarrow 16x_1^4 = 4y_1^2 + 4z_1^2 + 4 = 4(y_1^2 + z_1^2 + 1) \Longleftrightarrow 4x_1^2 = y_1^2 + z_1^2 + 1$$

The LHS is $0 \pmod 4$ while the RHS is $1, 3 \pmod 4$. Contradiction.

I don't know how to proceed with Case 2 when $y$ and $z$ are of unlike parity. Could you give me some hints?

Is my proof for the first case correct?

Could you suggest some better proofs?


  • $\begingroup$ The last RHS can also be $2\pmod 4$, if for example $y_1$ is even and $z_1$ is odd. This is a detail though. $\endgroup$ – Wojowu Mar 28 '16 at 10:11
  • $\begingroup$ No, because if $y,z$ are both even, then $y_1,z_1$ (which I assume they are $y/2,z/2$; if that's wrong, you should specify what they are in the question body) can have either parity. $\endgroup$ – Wojowu Mar 28 '16 at 10:14
  • $\begingroup$ @Wojowu Thanks for pointing out the error. $\endgroup$ – TheRandomGuy Mar 28 '16 at 10:17
  • $\begingroup$ The case of odd $y,\,z$ is "similar" in that it has no solutions either; other than that, it's hardly similar and should not be waived away like that. $\endgroup$ – Ivan Neretin Mar 28 '16 at 10:33
  • 1
    $\begingroup$ hint: see $\mod 8$ $\endgroup$ – Mikhail Ivanov Mar 28 '16 at 21:01

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