Difference in squares

Since $n$ and $m$ are perfect squares, then we know that the equation $n-m=63$ must have the form of $x^{2} - y^{2} = 63$ , or $(x-y)(x+y)=63$ .

$63$ can likewise be factored into $(1,63),(3,21),$ and $(7,9)$ .

Now, we can associate each factor on one side to a factor on the other.

$\begin{aligned} x-y &= 1 & x+y &= 63 \\ x-y &= 3 & x+y &= 21\\ x-y &= 7 & x+y &= 9\\ x-y &= 9 & x+y &= 7 \\ x-y &= 21 & x+y &= 3\\ x-y &= 63 & x+y &= 1\\ \end{aligned}$

The last three equations are identical to the first three so they can be ignored.

Solving for $x$ and $y$ , we get:

$\begin{aligned} x&=32 & y &= 31 \\ x&=12 & y &= 9 \\ x&=8 & y &= 1 \\ \end{aligned}$

Thus, we have three distinct pairs of $n$ and $m$ which solve this equation.