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.
 A: Another way, again assuming that $x$ must be an integer: We have
$$
a^2 = x+10,\qquad b^2=x-79
$$
so $a^2-b^2=(a+b)(a-b)=89$ and, since we're in the integers and 89 is prime, we have
$$
a+b=89, \qquad a-b = 1
$$
from which it follows that $a=45, b=44$. Thus
$$
a^2=45^2=2025=x+10
$$
so $x=2015$, which puts it squarely (ahem) in the region of PuzzleLand where the answers are part of the current date.
A: Assuming you are looking for positive integers...
Consecutive squares differ by consecutive odd amounts. The first differences to $\{0,1,4,9,16,\ldots\}$ are $\{1,3,5,7,\ldots\}$.
You have two squares that are $89$ apart. They may or may not be consecutive. But for some string of consecutive odd numbers, the sum must be $89$.
$$
\begin{align}
(2k+1)+\left(2(k+1)+1\right)\cdots+\left(2(k+h)+1\right)&=89\\
2k(h+1)+h(h+1)+h+1&=89\\
(2k+h+1)(h+1)&=89\\
\end{align}$$
Since $89$ is prime, either $h=0$ and $k=44$, or well, there is no other possibility.
So $k=44$ and there is only one consecutive odd number in our sequence, meaning the two squares must be consecutive squares. We must be dealing with the difference between $44^2$ and $45^2$. So $x=45^2-10$.
A: 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 $a_{n+1}=x+10$, $a_n = x-79$. $$\begin{align} \ a_{n+1}-a_n & = 2n+1 \\
& = \ (x+10)-(x-79) \\
& = \ 89 \\ \end {align}$$ $$n=44 \space \Rightarrow \space \begin{align}
\ a_{45} & =\ 45^2 \\ 
& =\ 2025 \\ 
& =\ x+10 \\ \end{align}$$ $$x=2015$$
