Let $ p = 4n + 1$ cousin. Prove that p$ | n ^ n - 1$
The only thing I developed was:
$1 = n ^ n (mod p)$
$1 = n (mod p)$
$4 = 4n (mod p)$
Maybe it has to do with Fermat's cousins
Let $ p = 4n + 1$ cousin. Prove that p$ | n ^ n - 1$
The only thing I developed was:
$1 = n ^ n (mod p)$
$1 = n (mod p)$
$4 = 4n (mod p)$
Maybe it has to do with Fermat's cousins
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