Each of X and Y is a distinct positive integer, such that:

- X and Y are relatively prime, and:
- Each of (X
^{2} – 12)/Y and (Y^{2 }– 12)/X is an integer.

Does there exist an infinite numbers of pairs (X,Y) satisfying the given conditions?

Give reasons for your answer.

Well, X or Y can be 1. This gives rise to 3 solutions (1,1), (1,11) and (11,1).

(Y^2 - 12)/X is an integer, so if X is divisible by 2, then so is Y.

Similarly, if X is divisible by 3 then so is Y.

But neither can be the case, because X and Y are relatively prime.

So X and Y cannot be divisible by 2 or 3.

If (Y^2 - 12)/X and (X^2 - 12)/Y are both integers, then so are their products.

In other words, (X^2*Y^2 -12X^2 - 12Y^2 + 144)/XY is an integer.

Then, subtract XY to get (-12X^2 - 12Y^2 + 144)/XY is an integer.

X and Y are not divisible by 2 or 3, so we can divide by -12 to get

(X^2 + Y^2 - 12)/XY is an integer.

Add 2 to get ((X+Y)^2 - 12)/XY is an integer

Or, subtract 2 to get ((X-Y)^2 - 12)/XY.

But none of that seems to help.

Also not obviously helpful: Since X and Y are odd, let X = 2A + 1, Y = 2B + 1.

That seems unfruitful, especially since A and B do not need to be relatively prime.

Out of ideas.