Factorials and Squares.

Rewrite the equation as $y^{2} - x^{2} = 10! \Longrightarrow (y - x)(y + x) = 10!$ .

So $(y - x)$ and $(y + x)$ must be co-factors of $10!$ . Also, since they both have the same parity, (i.e., they will be either both even or both odd, (as $y + x = (y - x) + 2x$ )), and since $10!$ is even, both $(y - x)$ and $(y + x)$ must be even.

Now $10! = 2^{8}*3^{4}*5^{2}*7$ . After distributing a factor of $2$ to each of $(y - x)$ and $(y + x)$ , we are left with $2^{6}*3^{4}*5^{2}*7$ . Ignoring the $7$ for the moment, we note that there are $7*5*3 = 105$ divisors of $2^{6}*3^{4}*5^{2}$ , each of which is of the form

$2^{a}*3^{b}*5^{c}$ , where $0 \le a \le 6, 0 \le b \le 4, 0 \le c \le 2$ .

For a given $a,b,c$ , form two numbers as follows:

(A) $2^{a+1}*3^{b}*5^{c}*7$ and (B) $2^{7-a}*3^{4-b}*5^{2-c}$ .

In this way we have formed distinct, even co-factors of $10!$ , (distinct because of the factor $7$ in (A)). As they are distinct, we can assign the lesser of the two to $(y - x)$ and the greater to $(y + x)$ . As they are both even we can then solve for a unique, ordered, positive integer pair $(x,y)$ .

As there are $105$ ways in which to construct the pair ((A), (B)), there will thus be $\boxed{105}$ ordered, positive integers pairs $(x,y)$ satisfying the given equation.

We can rewrite the expression as $(y+x)(y-x) = 2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7$ . Notice that both factors must be even because otherwise, after finding both factors, we would have $2y = even + odd$ so $y = odd/2$ and you know that then $y$ wouldn't be positive integer. So this shows that both $(y+x)$ and $(y-x)$ must be even, so we distribute one $2$ to each. Then we can distribute the prime factorization $2^{6} \cdot 3^{4} \cdot 5^{2} \cdot 7^{1}$ to both factors, but in how many ways? Well, we have $7\cdot5\cdot3\cdot2 = 210$ ways of distributing them (remember, we're distributing to the factors $(y+x)$ and $(y-x)$ ). But why this is not the answer, and is $105$ ? There are to important ordered pairs $((y+x), (y-x))$ , that we should see. The first one: When $y+x = 2^{3}\cdot 3^{2}\cdot5\cdot7$ , then $y-x = 2^{3}\cdot 3^{2}\cdot5$ and here we yet have $x, y > 0$ . But the second one: When $y+x = 2^{3}\cdot 3^{2}\cdot5$ , then $y-x = 2^{3}\cdot 3^{2}\cdot5\cdot7$ . Well, if we put the $210$ combinations in order of decreasing $y+x$ , this would be the transition from the $105^{th}$ ordered pair to the $106^{th}$ ordered pair. For the $106^{th}$ , notice that $y-x > y+x$ so $x < -x$ , and hence, $x \leq 0$ . So we have $\boxed{105}$ ordered pairs (x, y) that satisfy the equation.

PD.: I'm so excited, it has been a while since i dont enter to brilliant and i was able to make this problem in my mind :').