Problem
NT-B1-M07-P014 Divisibility for All \(k\)
#14
★★★☆☆ Level 3 of 5
Prove that \(13\mid 5^{12k}-1\) for every positive integer \(k\).
First apply Fermat to \(5^{12}\).
Since \(13\) is prime and \(13\nmid5\), \(5^{12}\equiv1\pmod{13}\). Then \(5^{12k}=(5^{12})^k\equiv1^k\equiv1\pmod{13}\). Hence \(13\mid5^{12k}-1\).
The problem teaches raising a congruence to a power.