Problem
NT-B2-M06-P014 Guaranteed Divisibility
#14
★★★★☆ Level 4 of 5
Let \(p\) be an odd prime, \(p\mid a-b\), \(p\nmid ab\). Prove that \(p^r\mid a^{p^{r-1}}-b^{p^{r-1}}\) for every \(r\ge1\).
Apply LTE to the exponent \(p^{r-1}\).
By LTE, \(v_p(a^{p^{r-1}}-b^{p^{r-1}})=v_p(a-b)+v_p(p^{r-1})\). Since \(p\mid a-b\), the first term is at least \(1\), and the second equals \(r-1\). The total exponent is at least \(r\), so divisibility by \(p^r\) follows.
Shows LTE as a proof tool.