How many pairs of positive integers $(x,y)$ satisfy $x^2-y^2=5100$?

Answer 1

The equation factors as $(x+y)(x-y)=5100 = 2^2\cdot3\cdot5^2\cdot17$. $x+y$ and $x-y$ must be either both even or both odd because their sum and their difference are both even. Prime factorizing $5100$ yields $2^2\cdot3\cdot5^2\cdot17$, so the prime factorization of $x+y$ and $x-y$ must be $2^a\cdot3^b\cdot5^c\cdot17^d$ and $2^{2-a}\cdot3^{1-b}\cdot5^{2-c}\cdot17^{1-d}$ where $a,b,c,d$ are all either $0$ or $2$. To get the total number of ways for the expression to be a factorization of $5100$, we have $2\cdot2\cdot2\cdot2 = 16$ choices (each can either be $0$ or $2$). However, we must divide by $2$ at the end because each pair will be counted twice. The answer is $\frac{16}{2} = \boxed{8}$.