Problem
ALG-B1-M02-P022 Two Differences of Squares
#22
★★★★☆ Level 4 of 5
Find all integer pairs \(x,y\) such that \(x^2-y^2=24\) and \(x-y>0\).
Set \(u=x-y\), \(v=x+y\); the numbers \(u\) and \(v\) have the same parity.
We have \((x-y)(x+y)=24\). Let \(u=x-y>0\), \(v=x+y\). Then \(uv=24\), and \(u\) and \(v\) have the same parity.
The suitable positive pairs are \((2,12)\) and \((4,6)\). From \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\), we get \((x,y)=(7,5)\), \((5,1)\).
If \(v<0\), then \(uv<0\) since \(u>0\), so there are no other solutions.
Parity must be tracked; otherwise half-integer false answers appear.