Difference of Squares (2)

If $2520 = x^2 - y^2$ then $2520 = (x+y)(x-y) = s\cdot d.$ The sum and difference must be positive factors of 2520. Moreover, both must be odd or both must be even. Since 2520 is even, both $s$ and $d$ must be even. Thus we write $s = 2s'$ and $d = 2d'$ ; and $s'\cdot d' = 630.$ Since 630 is no perfect square, we are guaranteed that $s' \not= d'$ . We also require $s' > d'$ . It follows that the set of all values $s', d'$ are precisely all the factors of 630.

Now $630 = 2\cdot 3^2\cdot 5\cdot 7;$ Its factors are of the form $2^a\cdot 3^b\cdot 5^c\cdot 7^d$ with $0 \leq a, c, d \leq 1$ and $0 \leq b \leq 2$ . There are $2 \times 3 \times 2 \times 2 = 24$ of such factors. Because this is the number of $s'$ and $d'$ values together, there are $24/2 = \boxed{12}$ pairs of values that satisfy.

$\begin{array}{cc|l} s' & d' & \\ \hline 630 & 1 & 631^2 - 629^2 = 2520 \\ 315 & 2 & 317^2 - 313^2 = 2520 \\ 210 & 3 & 213^2 - 207^2 = 2520 \\ 126 & 5 & 131^2 - 121^2 = 2520 \\ 105 & 6 & 111^2 - 99^2 = 2520 \\ 90 & 7 & 97^2 - 83^2 = 2520 \\ 70 & 9 & 79^2 - 61^2 = 2520 \\ 63 & 10 & 73^2 - 53^2 = 2520 \\ 45 & 14 & 59^2 - 31^2 = 2520 \\ 42 & 15 & 57^2 - 27^2 = 2520 \\ 35 & 18 & 53^2 - 17^2 = 2520 \\ 30 & 21 & 51^2 - 9^2 = 2520 \\ \hline \end{array}$

If $2520 = m^2-n^2$ , then $2520 = (m+n)(m-n)$

Here, $(m+n)$ and $(m-n)$ both should be either even or odd. As, $2520$ is even both of them will be even.

now, $2520 = 2^3*3^2*5*7 = 1*2520 = 2*1260 = 3*840 = 4*630 = 5* 504 = 6*420 = 7*360 = 8*315 = 9*280 = 10*252 = 12*210 = 14*180 = 15*168 = 18*140 = 20*126 = 21*120 = 24*105 = 28*90 = 30*84 = 35*72 = 36* 70 = 40*63 = 42*60 = 45*56$

So, we get a total of 12 pairs.

I think the answer rather depends on the domain of a and b which is unstated in the question? $\\$ If we consider $a,b \in \R \\$ Then $a=\pm1250.5$ and $b=\pm1249.5$ are also solutions yielding 32 possible expressions. $\\$ e.g. $(1250.5)^2 - (1249.5)^2$ and $(-1250.5)^2 - (1249.5)^2$ and $(1250.9)^2 - (-1249.5)^2$ and $(-1250.5)^2-(-1249.5)^2\\$ If we only consider $a,b \in \R^+$ (Positive Real numbers) Then I think there are 8 possible expressions. $\\$ If we consider $a,b \in \Z$ (Positive and negative integers) I think there are 12 possible expressions. $\\$ If we consider $a,b \in \N$ (Positive Integers) I think there are only 3 possible expressions.

Mathematica

Solve[a^2 - b^2 == 2520 && 0 < b < a, {a, b}, Integers]

{{a -> 51, b -> 9}, {a -> 53, b -> 17}, {a -> 57, b -> 27}, {a -> 59, b -> 31}, {a -> 73, b -> 53}, {a -> 79, b -> 61}, {a -> 97, b -> 83}, {a -> 111, b -> 99}, {a -> 131, b -> 121}, {a -> 213, b -> 207}, {a -> 317, b -> 313}, {a -> 631, b -> 629}}