True or False (1)

The statement is $\boxed{\text{False.}}$ To show this, we will prove the stronger [at least superficially] statement: $\text{There is no nonzero integral solution to }\, a^2 + b^2 + c^2 = a^2b^2,$ by repeatedly using the well-known fact $n^2 \equiv \begin{cases} 0 \pmod{4}, & \text{ if } n \text{ is even} \\ 1 \pmod{4}, & \text{ if } n \text{ is odd}\end{cases}.$

Suppose there is a nonzero integral solution $(a_0,b_0,c_0).$ Then the equation can be written as $c_0^2 + 1 = (1 - a_0^2)(1 - b_0^2)$ If $c_0$ is odd, then $c_0 \equiv 1 \pmod{4}$ and $1 - a_0^2, 1 - b_0^2 \equiv 0 \text{ or } 1 \pmod{4}$ so the equation becomes $2 \equiv c_0^2 + 1 = (1 - a_0^2)(1 - b_0^2) \equiv 0 \text{ or } 1 \pmod{4}$ which is a contradiction, so we can conclude $c_0$ is even. Therefore, using the same form of the equation mod $4$ again, we have $1 \equiv c_0^2 + 1 = (1 - a_0^2)(1 - b_0^2) \pmod{4} \implies a_0^2 \equiv 0 \equiv b_0^2 \pmod{4} \implies a_0,b_0 \text{ are even}$

Since each of $a_0,b_0,c_0$ is even and at least one is nonzero, there is some positive integer $k$ such that $2^k$ divides each of $a_0,b_0,c_0$ and at least one of the following is odd: $a_1 = \frac{a_0}{2^k}, \qquad b_1 = \frac{b_0}{2^k}, \qquad c_1 = \frac{c_0}{2^k}.$

Using these new values, the given equation becomes $a_1^2 + b_1^2 + c_1^2 = 4^k a_1^2 b_1^2$

Since $k$ is positive, $4^k \equiv 0 \pmod{4},$ so $a_1^2 + b_1^2 + c_1^2 = 4^k a_1^2 b_1^2 \equiv 0 \pmod{4} \implies a_1 \equiv b_1 \equiv c_1 \equiv 0 \pmod{2}$ but then none of $a_1, b_1, c_1$ is odd, contradicting the definition of $k.$

This final contradiction is only under the assumption that there is some nonzero integral solution, so we can conclude that there is no such solution. In particular, there is no solution using positive integers.