Answer
See proof.
Work Step by Step
Assume by contradiction that both $n$ and $n+2$ are perfect squares. It follows that $n=a^2$ and $n+2=b^2$ for some integers $a,b$. Notice now that $2=(n+2)-n=b^2-a^2=(b+a)(b-a)$ where $b+a$ and $b-a$ are integers. But, then we have that a product of two integers is equal to 2, a contradiction. Thus, if $n$ is a perfect square $n+2$ is not.
