# Prove that 61! = -1mod71

Note by Nicholas Fortino
5 years, 8 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 $$71$$).

- 5 years, 8 months ago

Oh I see why I was looking at it the wrong way. Nice proof Zi song :)

- 5 years, 8 months ago

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 ?

- 5 years, 8 months ago