Diophantine equation $a^2+b^2+c^2=a^2b^2$

Question

I am trying to find all non trivial integers for which $a^2+b^2+c^2=a^2b^2$. As suggested I have tried working (mod 4).

This is what I've gotten so far:
Squares can have a remainder of 0 or 1 (mod 4). So the rhs can be 0 or 1 (mod 4). It is 1 if and only if both $a^2$ and $b^2$ have remainder 1 (mod 4).

But if both $a^2$ and $b^2$ are 1 the lhs has remainder 2 or 3 (if $c^2$ also has remainder 1), a contradiction.

So $a^2,b^2$ and $c^2$ must be divisible by 4.
Is this reasoning correct? If I have made any mistakes please let me know.

Answer 1

Just work from parity.

If $a,b$ are both odd then $a^2+b^2\equiv 2 \pmod 4$ and $a^2b^2\equiv 1 \pmod 4$ This would mean that $c^2\equiv -1\pmod 4$ which is impossible.

If $a=2A$ and $b$ is odd then we get $1+c^2\equiv 0\pmod 4$ which is, again, impossible.

If $a,b$ are both even then $c$ is also even. Writing them as $2A,2B,2C$ we deduce that $$A^2+B^2+C^2=4A^2B^2$$

As before: if $A,B$ are both odd then we get $C^2\equiv 2\pmod 4$, which is impossible. If $A$ is even and $B$ is odd we get $C^2\equiv -1\pmod 4$ which is impossible. Hence, again, $A,B$ are both even. This repeats (with higher powers of $2$ appearing on the right) so infinite descent finishes the job.

Answer 2

One way is to prove via the infinite descent method (i.e., showing that $a$, $b$, and $c$ is divisible by $2^k$ for each $k=1,2,\ldots$). I am offering an alternative solution.

Without loss of generality, suppose that $a\neq 0$. It can be easily seen that $a$, $b$, and $c$ are even. Furthermore, we have $$\left(a^2-1\right)\left(b^2-1\right)=c^2+1\,.$$ Since $a^2-1$ is a natural number congruent to $3$ modulo $4$, it has a prime divisor $p\equiv 3\pmod{4}$. However, this means $c^2\equiv -1\pmod{p}$, and we know that $-1$ is not a quadratic residue modulo $p$.