 ## Diophantine Challenge

#### Problem

Given that x, y, and b are positive integers, prove that the Diophantine equation, x2 + (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, x2 1 = (x b)y.

When b=1, and taking the subtract form of LHS, we get x2 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, x2+a = (x b)y.

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

Similarly x b divides (a+bx) (bx b2) = b2+a.

So a solution exists for each value of x b that divides b2 1, or rather each factor of b2 1.

When x b = b2 1 or x b = b2+1, we get x = b2+b 1, which are both positive integers. And as we have already established that x b divides both sides of the Diophantine equation, y = (x2 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 b2+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 = x2 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.

Problem ID: 227 (04 Jun 2005)     Difficulty: 4 Star

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