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Prove that 61! = -1mod71

Note by Nicholas Fortino
4 years, 7 months ago

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This is a simple application of Wilson's Theorem.

First, note that \(9! \equiv -1\) (mod \(71\)).

Then, since \(71\) is a prime, by Wilson's Theorem,

\(70! \equiv -1\) (mod \(71\)).

But \(70! \equiv 61! (62)(63)(64)...(70) \equiv 61! (-9)(-8)(-7)...(-1)\) \( \equiv -61!(9!) \equiv 61! \) (mod \(71\)).

But \(70! \equiv -1\) (mod \(71\)).

So \(61! \equiv 70! \equiv -1\) (mod \(71\)). Zi Song Yeoh · 4 years, 7 months ago

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Oh I see why I was looking at it the wrong way. Nice proof Zi song :) Johnson Adeleke · 4 years, 7 months ago

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I'm not that great at modulus arithmetic but how can 61!= -1 mod(7) ? Doesn't that mean that it is smaller than 7 ? Johnson Adeleke · 4 years, 7 months ago

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