Math puzzles

Originally posted by QuickDrawd4
It certainly is a problem...

First we can state the problem as:
(p-1)!+1 = p^k for some k.

assume k=1.

(p-1)!+1=p
=> (p-1)!=p-1

which has only the trivial solution of p=2.

Now assume k=2
(p-1)!+1=p^2

Primes for which (p-1)!+1 = 0 mod p^2 are called Wilson Primes, only 3 are known:
5,13 and 563 with no others below 5*10^8
source: ht ...[text shortened]... with p < 5*10^8.

It is conjectured however that there are an infinite number of Wilson primes

i like the analysis... though you did miss the other trivial solution where p=3 🙂

interesting to note though that (p-1)! = (p-1)(p-2)(p-3)...(3)(2)(1) which is divisible by every prime less than p. but then (p-1)! + 1 is not divisible by ANY primes less than p, and as such must be divisible by p or a prime larger than p, or a product of primes greater than or equal to p. (reminiscent of the proof of the infinitude of the primes)

by cursory inspection, i believe (p-1)! = -1 mod p but i don't remember the theorem/proof... (is it related to fermat's little theorem?) and as such i think that [(p-1)! + 1] is always a MULTIPLE of p, though this is not what the op was asking 🙂 also i'm too lazy to support this with calculations and incidentally, may be completely wrong!

Originally posted by QuickDrawd4
It certainly is a problem...

First we can state the problem as:
(p-1)!+1 = p^k for some k.

assume k=1.

(p-1)!+1=p
=> (p-1)!=p-1

which has only the trivial solution of p=2.

Now assume k=2
(p-1)!+1=p^2

Primes for which (p-1)!+1 = 0 mod p^2 are called Wilson Primes, only 3 are known:
5,13 and 563 with no others below 5*10^8
source: ht ...[text shortened]... with p < 5*10^8.

It is conjectured however that there are an infinite number of Wilson primes

(p-1)!+1=p
=> (p-1)!=p-1

which has only the trivial solution of p=2.

This is incorrect, which led to your overlooking p=3 as a solution.

Otherwise, you are basically correct that the only solutions are p=2,3,5.

That there are no solutions above p=5 can be proven without knowing anything about the subject of Wilson primes (for instance, using proof by contradiction). Which is not to say that it's not still a tricky problem.

Originally posted by Aetherael
i like the analysis... though you did miss the other trivial solution where p=3 🙂

interesting to note though that (p-1)! = (p-1)(p-2)(p-3)...(3)(2)(1) which is divisible by every prime less than p. but then (p-1)! + 1 is not divisible by ANY primes less than p, and as such must be divisible by p or a prime larger than p, or a product of primes great lso i'm too lazy to support this with calculations and incidentally, may be completely wrong!

by cursory inspection, i believe (p-1)! = -1 mod p but i don't remember the theorem/proof... (is it related to fermat's little theorem?)

Correct, it's Wilson's Theorem. Similar to Fermat's little theorem (but Wilson's is necessary and sufficient for primality, whereas Fermat's isn't).