Are there any integers $a,b$ s.t. ${ a }^{ 2 }-{ b }^{ 2 }=8$? For $a$ and $b$ are integers greater than $1$, ${ a }^{ 2 }-{ b }^{ 2 }=8$ holds? 
 A: Hint: We have $8= a^2 - b^2 = (a+b)(a-b)$. As $a,b \ge 1$, we have $a+b > a-b$, that is either $a+b=8$ and $a-b = 1$, or $a+b=4$ and $a-b=2$.
A: For that to be true, 
$$
(a-b)(a+b) = 8,
$$
where a and b are integers,
so $a-b = 1$ and $a+b=8$ or $a-b=2$ and $a+b=4$. Note that the sum must be the bigger one. The second system has integer solutions, but $a=3$ and $b=1$. So the answer is no. 
A: Write $(a-b)(a+b) = 8$. You have only four different posibilities
$a+b = \pm 2$ or $a-b = \pm 4$ and the other way around.
Note that $a+b$ is even iff $a-b$ is even.
These all have unique solution in $\mathbb{Q}$, because of a linear algebra argument.
Now, you have to check whether you get an integer $>1$ for any of them.
A: If you allow $1$, then $a=3$ and $b=1$ are a solution.
Else, notice that you must have $a>b$, write $a = b+n$. Then
$$a^2 - b^2 = 2bn+n^2\stackrel{!}{=} 8$$
Since $n,b>0$, we must have $8-n^2>0$, and thus $n=1$ or $n=2$.


*

*$n=1$: $2b+1 = 8 \Leftrightarrow 2b=7$ and this is impossible.

*$n=2$: $4b+4=8\Leftrightarrow b=1$ and that's the solution mentioned at the beginning of my answer.


In conclusion, the only possible solution for $a,b\geq0$ is $a=3$, $b=1$.
