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Using Fermat's Little Theorem, prove $1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv -1 \pmod p$ where $p$ is a prime.

So we would use Fermat's theorem: $a^{p-1}\equiv 1\pmod p$

Would the proof go something like this....?

$1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv 1+\ldots+1$

There would be $p-1$ many $1$'s so (i.e. $p-1=5-1=4$ so $1+1+1+1=4$)

$p-1 \equiv -1\pmod p$

$p \equiv 0 \pmod p$

Would that be right?

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    The fun begins when you try to prove the converse: if the congruence holds, then $p$ is prime. This is known as *Giuga's conjecture*, proposed in 1950, verified up to enormous values of $p$, but not yet proved.2012-12-10

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