mathschallenge.net

Problem

Given that x, y, and b are positive integers, prove that the Diophantine equation, x² + (bx)y = 1 in x and y, has at least four solutions for all values of b.

Solution

We shall begin by rearranging the equation, x²1 = (xb)y.

When b=1, and taking the subtract form of LHS, we get x²1 = (x+1)(x1) = (x1)y, so y = x+1. That is, we have infinitely many solutions for (x,y): (1,2), (2,3), (3,4), ... .

For b2, let us deal with a slightly more general form, x²+a = (xb)y.

Clearly xb divides x²bx, and as xb divides the RHS, it follows that it must divide x²+a. Therefore, xb divides (x²+a)(x²bx) = a+bx.

Similarly xb divides (a+bx)(bxb²) = b²+a.

So a solution exists for each value of xb that divides b²1, or rather each factor of b²1.

When xb = b²1 or xb = b²+1, we get x = b²+b1, which are both positive integers. And as we have already established that xb divides both sides of the Diophantine equation, y = (x²1)/(xb) will also be positive integers. Thus we have two positive integer solutions for x and y.

But when xb = 1, we can see that x = b+1 b²+b1 for b2, and so this solution in x will be different to the previous two. In addition, by substituting xb = 1 into the Diophantine equation, we get y = x²1, 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.