# Difference between revisions of "2006 IMO Problems/Problem 4"

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If <math>(x,y)</math> is a solution then obviously <math>x\geq 0</math> and <math>(x,-y)</math> is a solution too. For <math>x=0</math> we get the two solutions <math>(0,2)</math> and <math>(0,-2)</math>. | If <math>(x,y)</math> is a solution then obviously <math>x\geq 0</math> and <math>(x,-y)</math> is a solution too. For <math>x=0</math> we get the two solutions <math>(0,2)</math> and <math>(0,-2)</math>. | ||

## Latest revision as of 13:59, 31 July 2019

### Problem

Determine all pairs of integers such that

### Solution

If is a solution then obviously and is a solution too. For we get the two solutions and .

Now let be a solution with ; without loss of generality confine attention to . The equation rewritten as shows that the factors and are even, exactly one of them divisible by . Hence and one of these factors is divisible by but not by . So Plugging this into the original equation we obtain or, equivalently Therefore For this yields , i.e. , which fails to satisfy . For equation gives us implying . Hence ; on the other hand cannot be by . Because is odd, we obtain , leading to . From we get . These values indeed satisfy the given equation. Recall that then is also good. Thus we have the complete list of solutions : , , , .