Prove that there are infinitely many ordered triples of positive integers $\displaystyle (a, b, c)$ such that $\displaystyle gcd\ (a, b, c) = 1$ and $\displaystyle a^2b^2 + b^2c^2 + c^2a^2$ is a perfect square.
put $\displaystyle a=b=1.$ then you only need to find infinitely many $\displaystyle c \in \mathbb{N}$ such that $\displaystyle 2c^2 + 1 = x^2,$ for some $\displaystyle x \in \mathbb{N}.$ the equation $\displaystyle x^2 - 2c^2=1$ is known as Pell equation. it obviously has a solution
$\displaystyle x=3, \ c = 2.$ the general solution for $\displaystyle c$ is known to be: $\displaystyle c=\frac{(3+2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{2\sqrt{2}}, \ \ n \in \mathbb{N}. \ \ \ \Box$
Note: it's a good exercise to prove directly, without using what is known about Pell equation, that $\displaystyle c_n=\frac{(3+2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{2\sqrt{2}} \in \mathbb{N},$ for all $\displaystyle n \in \mathbb{N},$ and also that $\displaystyle 2c_n^2 + 1$ is a perfect square.