Prove that the only integer solution of the equation $a^2 + b^2 + c^2 = a^2 b^2$ is $a = b = c = 0$.

Question

Any tips are welcome! I have tried various methods, but to no avail. I could use a pointer as to the right direction. Thank you.

Answer 1

Note that the given equation is equivalent to $$(a^2-1)(b^2-1)=c^2+1\tag{*}$$ Then consider two cases:

1) If $a$ or $b$ is odd then $a^2\equiv 1$ or $b^2\equiv 1$ modulo $4$ and (*) implies that $c^2\equiv -1$ modulo $4$ which is impossible.

2) If $a$ and $b$ is even then $a^2\equiv 0$ or $b^2\equiv 0$ modulo $4$. By (*) it follows that $c^2\equiv 0$ modulo $4$ that is $c$ is even. Now if $a=2A$, $b=2B$ and $c=2C$, then $A^2+B^2+C^2\equiv 0$ modulo $4$ and again $A$, $B$, $C$ are even and, by infinite descent, there is a unique solution $(0,0,0)$.

Answer 2

Recall that squares of integers are $\equiv 0$ or $1\pmod 4$, depending on the parity.

Consider $$\tag1a^2+b^2+c^2=(2^kab)^2$$ with integers $a,b,c$ and non-negative integer $k$. Of course, this has the trivial solutions $(a,b,c,k)=(0,0,0,k)$, $k$ arbitrary. Assume there are also non-trivial solutions and among these pick one that minimizes the natural number $|a|+|b|+|c|$.

If the square on right hand side is $\equiv 1\pmod 4$, then $k=0$ and $a,b$ must both be odd. Then the left hand side is $\equiv 2$ or $3\pmod 4$, contradicton. Hence the the square on the right is $\equiv 0\pmod 4$. The left hand side is $\equiv 0,1,2,3\pmod 4$, depending on how many of $a,b,c$ are odd. We conclude that $a,b,c$ are even and then $(\frac a2,\frac b2,\frac c2,k+1)$ is also a solution.

This contradicts the minimality of the solution we started with.