$x^4-4=y^2+z^2$ prove that it has no integer solution $x^4-4=y^2+z^2$ prove that it has no integer solution
I tried to check mod$4$  , mod $3$ ...
It doesn't give anything.
I want to solve this problem
by supposing  that I'm finding the smallest solution and then prove that a smaller one exists. But for that I have to find something on which every member of equation divides.Which I couldn't find.
 A: Outline/hints: first show that $x$ cannot be even by looking modulo $16$. Then, for every odd prime $p$ dividing $x^4-4$:

*

*Since $p$ divides $y^2+z^2$, we must have $p\equiv 1\pmod 4$.

*Since $x^4\equiv2^2\pmod p$, show that $2$ is a quadratic residue modulo $p$, and conclude that $p\equiv 1\pmod 8$.

But this implies that $x^4-4\equiv1\pmod 8$, which will be a contradiction.
A: Let $A=x^2+2,B=x^2-2$. Then the equation can be written as $$AB=y^2+z^2.$$ Clearly $\gcd(A,B)|(A-B)=4$. One has two cases to consider.
Case 1. $2|\gcd(A,B)$.
In this case, $x=2x’$ is even and $y,z$ must have the same parity. The case when $y,z$ are both odd can be ruled out, otherwise, taking congruence mod 4 would yield a contradiction. Writing $y=2y’,z=2z’$, one has then $$(4x’^2+2)(4x’^2-2)=4(y’^2+z’^2)$$ $$\Rightarrow 4x’^4-1=y’^2+z’^2$$ which is impossible since a sum of two squares is not congruent to 3 mod 4.
Case 2. $\gcd(A,B)=1$.
In this case, $x$ must be odd. But since $\gcd(A,B)=1$ and $AB$ is a sum of two squares, any prime factor $p$ in $A$ must be either congruent to 1 mod 4 or congruent to 3 mod 4 with even multiplicity. It follows that $A$ must be congruent to 1 mod 4. But since $x$ is odd in this case, $A=x^2+2$ is congruent to 3 mod 4, which is a contradiction.
QED
