Discrete mathematics - Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$. Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.
This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
 A: Here are some hints. 
Case $1$: $a$ or $b$ is even. 
Then RHS $\equiv 0 \pmod 4$. Then we get that the LHS is also $0 \pmod 4$. The LHS will be a sum of one square or two squares modulo $4$ in this case. This forces all of $a,b,c$ to be even since the quadratic residues modulo $4$ are either $0$ or $1$. Thus, unless all $a,b,c$ are $0$, we can divide out the powers of $2$ to get to case $2$.
Case $2$: Both $a,b$ are odd.
Then RHS is $1 \pmod 4$ and LHS is $2+c^2 \pmod 4$. This equation is impossible.
Thus the only solution is $(0,0,0)$.
A: The equation is equivalent to
$$
c^2+1=(a^2-1)(b^2-1)
$$
Since the left side is either $1$ or $2$ mod $4$, and the right side is either $0$ or $1$ mod $4$, they must both be $1$ mod $4$, which means that $a,b,c$ must be even.
Since any prime which is $3$ mod $4$ is a Gaussian prime, no prime which is $3$ mod $4$ can divide $c^2+1=(c+i)(c-i)$. Thus, the only primes which can divide $c^2+1$ must be $1$ mod $4$.
Since $a^2-1$ and $b^2-1$ are both $3$ mod $4$, they must have a prime factor which is $3$ mod $4$ and hence cannot divide the left hand side (except if they are both $-1$). Thus, the only solution is when $a,b,c=0$.
