How many pairs of positive integers $(x,y)$ satisfy $x^2-y^2=5100$?
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 $\frac{16}{2} = \boxed{8}$.
To find the pairs of positive integers $(x, y)$ that satisfy the equation $x^2 - y^2 = 5100$, we can rewrite this equation as $(x+y)(x-y) = 5100$.
We are looking for pairs of positive integers that multiply to give 5100. We can find the prime factorization of 5100 to determine all the possible pairs $(x+y, x-y)$.
The prime factorization of 5100 is $2^2 \times 3 \times 5^2 \times 17$. Now, we need to find all possible pairs of factors that multiply to give $5100$.
The factors of $5100$ (excluding $1$ and $5100$) are:
$2, 3, 4, 5, 6, 10, 12, 15, 17, 20, 25, 30, 34, 50, 51, 68, 75, 85, 100, 102, 136, 150, 170, 204, 255, 300, 340, 425, 510, 600, 680, 850, 1020, 1275, 1700, 2040, 2550, 3400, 5100$
Now, we need to find the pairs $(x+y, x-y)$ that match the factors of $5100$.
Taking each factor $a$ from the list, we solve the system of equations:
$x + y = a$
$x - y = \frac{5100}{a}$
Adding the two equations gives:
$2x = a + \frac{5100}{a}$
$x = \frac{a}{2} + \frac{2550}{a}$
Now, we check if $a$ and $\frac{5100}{a}$ have the same parity (both even or both odd). If they have the same parity, then the values of $x$ and $y$ will be positive integers. Otherwise, there are no positive integer solutions.
Now, let's go through the list of factors.
For $a = 2$, $\frac{5100}{a} = 2550$ and they have the same parity. So we have:
$x = \frac{2}{2} + \frac{2550}{2} = 1 + 1275 = 1276$
$y = \frac{2550}{2} - \frac{2}{2} = 1275 - 1 = 1274$
For $a = 3$, $\frac{5100}{a} = 1700$ and they have the same parity. So we have:
$x = \frac{3}{2} + \frac{2550}{3} = \frac{3}{2} + 850 = 851.5$
$y = \frac{2550}{3} - \frac{3}{2} = 850 - 1.5 = 848.5$
Since $x$ and $y$ are not positive integers, the pair $(x, y) = (851.5, 848.5)$ does not satisfy the conditions.
Continuing this process for all other factors, we find that the only pair of positive integers that satisfies the equation $x^2 - y^2 = 5100$ is $(x, y) = (1276, 1274)$.