#### numbthenoob

Is there a name and/or proof for the following conjecture?

"For any prime p, p! is congruent to p2-p modulo p2."

Thanks much.

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#### Dickfore

I think this might be of interest:

"[URL [Broken]

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#### numbthenoob

Thanks, Dickfore, that was of interest, even if most of it was over my head.

However, if I understand correctly...Euler's theorem is about the relationship between coprimes. I'm researching numbers that are not coprime, for example 7! and 72, where the gcd is 7, not 1.

To take another example, can I say with certainty that 66797! is congruent to 66797*66796 mod 667972 without having to calculate 66797!?

I've noticed that the pattern holds at least up to 13, i.e.

2! = 2 mod 4
3! = 6 mod 9
5! = 20 mod 25
7! = 42 mod 49
11! = 110 mod 121
13! = 156 mod 169

My question is: ...and so on? And a follow-up: if so, why?

#### Office_Shredder

Staff Emeritus
Gold Member
It might help to reduce the problem

p!=p2-p (mod p2) means that p2 divides (p!-p2+p)=p((p-1)!-p+1). We can divide a p out from everything

So really the question is, why is (p-1)!=p-1 (mod p). If you know that Zp is a field you can figure this out. If not, you probably still can, I just don't see how off the top of my head

#### Gib Z

Homework Helper
"[URL [Broken] Theorem[/URL]

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#### numbthenoob

Thanks Office Shredder and Gib Z, those were both very enlightening comments.

#### aprillove20

Good points Office_Shredder I learned something new formula.

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