Question

\(\large p\) is a prime number such that there exist positive integers \(\large n\) and \(\large m\) that satisfy the following equation:
\(\Large p^n+144=m^2\)
What is the sum of all possible values of \(\large m\) ?

Answer 1

\(\huge Solution\)
Rewriting gives: \(\Large p^n=(m+12)(m-12)\). Let \(\Large m+12=p^i\) and \(\Large m-12=p^j\). So \(\Large i+j=n\) and \(\Large i\gt j\)

Then we have:
\(\Large p^j(p^{i-j}-1)=p^i-p^j=(m+12)-(m-12)=24\\\Large =5^0(5^2-1)=2^3(2^2-1)=3^1(3^2-1)\)

From here we have 3 solutions: \(\Large m=13,15,20\) with sum \(\huge 48\)

Answer 2

Great work @JungHyunRan :)

Answer 3

thanks :)

Answer 4

brilliant ! xD