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Diophantine Challenge

Problem

Given that x, y, and b are positive integers, prove that the Diophantine equation, x2 + (bminusx)y = plus or minus1 in x and y, has at least four solutions for all values of b.


Solution

We shall begin by rearranging the equation, x2plus or minus1 = (xminusb)y.

When b=1, and taking the subtract form of LHS, we get x2minus1 = (x+1)(xminus1) = (xminus1)y, so y = x+1. That is, we have infinitely many solutions for (x,y): (1,2), (2,3), (3,4), ... .

For bgreater than or equal2, let us deal with a slightly more general form, x2+a = (xminusb)y.

Clearly xminusb divides x2minusbx, and as xminusb divides the RHS, it follows that it must divide x2+a. Therefore, xminusb divides (x2+a)minus(x2minusbx) = a+bx.

Similarly xminusb divides (a+bx)minus(bxminusb2) = b2+a.

So a solution exists for each value of xminusb that divides b2plus or minus1, or rather each factor of b2plus or minus1.

When xminusb = b2minus1 or xminusb = b2+1, we get x = b2+bplus or minus1, which are both positive integers. And as we have already established that xminusb divides both sides of the Diophantine equation, y = (x2plus or minus1)/(xminusb) will also be positive integers. Thus we have two positive integer solutions for x and y.

But when xminusb = 1, we can see that x = b+1 less than b2+bplus or minus1 for bgreater than or equal2, and so this solution in x will be different to the previous two. In addition, by substituting xminusb = 1 into the Diophantine equation, we get y = x2plus or minus1, 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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