Problem - 171
Let positive integer $d$ is a divisor of $2n^2$, where $n$ is also a positive integer. Prove $(n^2 + d)$ cannot be a perfect square.
If this does not hold, let $2n^2 = kd$ where $k$ is a positive integer. If $(n^2+d)$ is a perfect square, let it be $x^2$. We then have $$k^2x^2 = k^2(n^2 + d) = k^2n^2 + k^2d = k^2n^2 + k\cdot 2n^2 = n^2(k^2+2k)$$
Because $k^2 < k^2+2k < (k+1)^2$, therefore $(k^2+2k)$ cannot be a perfect square. This is a contradiction.