There's no other way to get 100!

By the equation, we have $x² - y² = 100 \Rightarrow (x+y)(x-y )= 100$ .

Since $x+y > x-y$ for $(x,y)$ are positive, then we have $(x+y) = (100, 50, 25, 20)$ and $(x-y) = (1,2,4,5)$ , for $(x,y)$ are positive integers.

If $x+y =100 \space ; \space x-y =1$ , there's no solution for $(x,y)$ .

If $x+ y = 50 \space ; \space x-y=2$ , we have $(x,y) = (26,24)$ .

If $x+y = 25 \space ; \space x-y = 4$ , there's no solution for $(x,y)$ .

If $x+y = 20 \space ; \space x-y = 5$ , there's no solution for $(x,y)$ .

Hence, there's only $\boxed{1}$ pair of positive integers of $(x,y)$ .

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By Triple Pythagoras, we can clearly see that $(x,y) = (26,24)$ .

$\begin{aligned} x^2 & = y^2 + 100 \\ x^2 - y^2 & = 100 \\ (x-y)(x+y) & = 100 \end{aligned}$

Let $mn = 100$ , where $m$ and $n$ be the positive integer factors of 100, with $m < n$ . Then we have:

$\begin{cases} x - y = m \\ x + y = n \end{cases} \implies 2x = m+n$ . Note that $m+n$ must be even so that $x$ is an integer. There are only four cases of $(m,n)$ :

$\begin{array} {cccl} m & n & m + n \\ 1 & 100 & 101 & \small \color{#D61F06} \text{Odd, rejected.} \\ 2 & 50 & 52 & \small \color{#3D99F6} \text{Even, }x = 26, \ y = 24 \\ 4 & 25 & 29 & \small \color{#D61F06} \text{Odd, rejected.} \\ 5 & 20 & 25 & \small \color{#D61F06} \text{Odd, rejected.} \end{array}$

Therefore, there is only $\boxed{1}$ case, that is when $x=26$ and $y=24$ .

Note that when $m=n=10$ , $\implies x = 10$ and $y=0$ , which is not a positive integer.