When is this congruence true for integer $a>1$?
$$(a-1)^{(a+1)} \equiv (a-1) \> mod \> (a^2-1)$$
My try:
$$(a-1)^{a} \equiv \> 1 \> mod \> (a+1)(a-1)$$
I do not know what to do now.
Thank you.
2026-03-25 12:23:57.1774441437
Noob question about an exponential congruence
30 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in MODULAR-ARITHMETIC
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