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?