If both $x+10$ and $x-79$ are perfect squares, what is $x$? 
If both $x+10$ and $x-79$ are perfect squares, what is $x$?

That is: 
$x+10=$perfect square  (or can be square such that the result of it squared is a positive integer)
$x-79=$perfect square  (or can be square such that the result of it squared is a positive integer)
I have no idea how to solve this.I tried inequality like
$$x+10>x-79$$
$$x-x>-79-10$$
$$0>-89$$
Note: You are not allowed to use trial and error or guess and check to solve the question.

It is helpful to keep in mind that the difference between consecutive perfect square numbers gives us an arithmetic sequence. $$\left\{n^2\right\}_{n=1} ^{\infty},\space \space \space a_n = n^2  $$ $$a_{n+1}-a_n =(n+1)^2-n^2= 2n+1$$ Therefore, we have $x+10$ and $x-79$ as consecutive perfect square numbers. Assume that $a_{n+1}=x+10$, $a_n = x-79$. $$\begin{align} \ a_{n+1}-a_n & = 2n+1 \\
& = \ (x+10)-(x-79) \\
& = \ 89 \\ \end {align}$$ $$n=44 \space \Rightarrow \space \begin{align}
\ a_{45} & =\ 45^2 \\ 
& =\ 2025 \\ 
& =\ x+10 \\ \end{align}$$ $$x=2015$$
