Problem
NT-B2-M07-P005 Difference of Squares of Primes
#5
★★☆☆☆ Level 2 of 5
Find all primes \(p>q\) such that \(p^2-q^2=120\).
Factor \((p-q)(p+q)=120\) and use the fact that odd primes give even factors.
If \(q=2\), then \(p^2=124\), impossible. Thus \(p,q\) are odd, and \(p-q\), \(p+q\) are positive even factors of \(120\). Check \((2,60),(4,30),(6,20),(10,12)\). These give \((p,q)=(31,29),(17,13),(13,7),(11,1)\). The last one is invalid because \(1\) is not prime. The answer is \((31,29),(17,13),(13,7)\).
It is useful to ask why the case \(q=2\) must be treated separately.