v8archie

As pointed out above, if you know a pair of factors $m$ and $n$ (with $m \gt n$), of your original number $k=mn$, you write $m=a+b$ and $n=a-b$. You can then solve for $a$ and $b$ $$a=\frac{m+n}2, \; b=\frac{m-n}2$$ and you have $$k=a^2-b^2$$

It's easy to see that, for integer solutions, you need $m$ and $n$ either both odd or both even. This isn't a problem for most numbers as for odd $k$ you may take $(m,n)=(k,1)$ (other pairs may exist); while for $k$ divisible by 4, $(m,n)=(\frac k 2,2)$ works (again other pairs may exist).

Even numbers not divisible by 4 won't have solutions.