Originally posted by QuickDrawd4i like the analysis... though you did miss the other trivial solution where p=3 🙂
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
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(p-1)!+1=p
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)!=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.
1 edit
Originally posted by Aetheraelby 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?)
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!
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).