$a^2-b^2 = x$ where $a,b,x$ are natural numbers Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.
Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?
Also, would there be a way to know how many such $(a,b)$ exists?

Note that $a^2-b^2=(a+b)(a-b)$ and that $(a+b)=c$ and $(a-b)=d$ are either both odd or both even.
If $x=cd$ where $c>d$ and $c$ and $d$ have the same parity then $a=\frac {c+d}2, b=\frac {c-d}2$ are natural numbers which work for $x$.
It is always possible to express an odd number $x$ in this way, using $c=x, d=1$. If $x$ is prime there is no other factorisation.
If $x$ is even, it must be divisible by 4, and if $x=4y$ then $c=2y, d=2$ will do provided $y>1$.
The number of solutions depends on the number of factorisations into factors of like parity.
