Problem
ALG-B1-M09-P007 Difference of squares
#7
★★☆☆☆ Level 2 of 5
Find all integer solutions of \(x^2-y^2=15\).
Factor into \(x-y\) and \(x+y\); the order of the factors matters too.
Let \(u=x-y\), \(v=x+y\). Then \(uv=15\), \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\). Both factors are odd, so all divisor pairs give integer \(x,y\).
From \((u,v)=(1,15),(15,1),(3,5),(5,3)\) and the corresponding negative pairs, we get \((8,7),(8,-7),(4,1),(4,-1),(-8,-7),(-8,7),(-4,-1),(-4,1)\).
It is important not to lose solutions with negative \(y\): the factor pairs \((u,v)\) and \((v,u)\) give different solutions.