Show that the only positive integer solutions for $a$ and $b$ in the equation $a^2-b^2=16$ are $a=4, b=0$ and $a=5, b=3.$ How many pairs of solutions would there be if we allowed negative values for the variables as well?
I know that $a^2-b^2=16=(a+b)(a-b)=16.$ I also know that the solutions are $(4,0)$ and $(5,3)$ because I plugged them in to find a solution. I just forgot how do you solve something like this. Can someone please show me? I suppose it is the samething with the negative values?