Problem
NT-B2-M02-P014 Minus One Modulo \(65\)
#14
★★★★☆ Level 4 of 5
Solve the congruence \(x^2\equiv-1\pmod{65}\).
Use \(65=5\cdot13\).
We need to solve \(x^2\equiv-1\pmod5\) and \(x^2\equiv-1\pmod{13}\). Modulo \(5\): \(x\equiv\pm2\). Modulo \(13\): \(x\equiv\pm5\). Combining the four pairs by CRT gives \(x\equiv8,18,47,57\pmod{65}\).
This connects quadratic residues with CRT without introducing the Legendre symbol.