Let $x$ be a positive real number, and $\lfloor{x}\rfloor$ be the largest integer that not exceeding $x$. Prove that there exist infinity number of positive integers, $n$, such that $\lfloor{\sqrt{2}}\ n\rfloor$ is a perfect square.
Because $(\sqrt{2}n)$ cannot be a square, this problem is equivalent to finding a positive integer, $x$, such that $x^2 < \sqrt{2} \ n < x^2+1$. If such $x$ exists, then $\lfloor{\sqrt{2} \ n}\rfloor = x^2$.
Consider the Pell's equation $x^2-2y^2=-1$. Because it has one solution $(1, 1)$, it has infinite number of solutions.
Multiplying $x^2$ on both sides of the equation and rearranging the terms leads to: $2x^2y^2 = x^4 + x^2$.
Therefore: $$x^4 < 2x^2y^2 < (x+1)^4 \implies x^2 < \sqrt{2} \ xy < (x+1)^2$$ Let $n=xy$, we then have $x^2 < \sqrt{2}\ n < (x+1)^2$. This means that there exists infinite number of such $x$.