## Diophantine Challenge

#### Problem

Given that `x`, `y`, and `b` are positive integers, prove that the Diophantine equation, `x`^{2} + (`b``x`)`y` = 1 in `x` and `y`, has at least four solutions for all values of `b`.

#### Solution

We shall begin by rearranging the equation, `x`^{2}1 = (`x``b`)`y`.

When `b`=1, and taking the subtract form of LHS, we get `x`^{2}1 = (`x`+1)(`x`1) = (`x`1)`y`, so `y` = `x`+1. That is, we have infinitely many solutions for (`x`,`y`): (1,2), (2,3), (3,4), ... .

For `b`2, let us deal with a slightly more general form, `x`^{2}+`a` = (`x``b`)`y`.

Clearly `x``b` divides `x`^{2}`bx`, and as `x``b` divides the RHS, it follows that it must divide `x`^{2}+`a`. Therefore, `x``b` divides (`x`^{2}+`a`)(`x`^{2}`bx`) = `a`+`bx`.

Similarly `x``b` divides (`a`+`bx`)(`bx``b`^{2}) = `b`^{2}+`a`.

So a solution exists for each value of `x``b` that divides `b`^{2}1, or rather each factor of `b`^{2}1.

When `x``b` = `b`^{2}1 or `x``b` = `b`^{2}+1, we get `x` = `b`^{2}+`b`1, which are both positive integers. And as we have already established that `x``b` divides both sides of the Diophantine equation, `y` = (`x`^{2}1)/(`x``b`) will also be positive integers. Thus we have two positive integer solutions for `x` and `y`.

But when `x``b` = 1, we can see that `x` = `b`+1 `b`^{2}+`b`1 for `b`2, and so this solution in `x` will be different to the previous two. In addition, by substituting `x``b` = 1 into the Diophantine equation, we get `y` = `x`^{2}1, which provides two more positive integer solutions for `x` and `y`.

Hence we have proved that the Diophantine equation has at least four positive integer solutions all values of `b`.

Prove that `b`=2 is the only value of `b` for which there are exactly four solutions.