second time puzzled.

We still need a proof for case p > 7 to show that for prime p

(p-1)! + 1 = p^a ------(1)

is only possible for p <=5.

In my last post , we have an estimate greatest power of 2 dividing (p-1)!.

f(p) >= (p - 1 - 2^n)/2 + (2^n -1 )


We must also have for (1) to hold that

p^a = 1 `mod` 2^f(p)

Our efforts will be to try to get a lower bound on a and if that proves to be too large (> p) for (p-1)!+1 to be equal to p^a then we are done.

We also know from Euler that

p^2^(f(p)-1) = 1 `mod` 2^f(p) since EulerPhi(2^(f(p)) = 2^(f(p)-1)

The smallest power s on p that leaves 1 as remainder mod 2^f(p)
should divide 2^(f(p)-1). Which means s is a power of 2. Most importantly, s should divide 'a' as well. All we need to prove then is that s becomes too large for p > 7.

So why is 's' special?
If 2^r >= s,
p^2^r = 1 `mod` 2^(f(p))

and if 2^r < s then
p^2^r /= 1 `mod` 2^(f(p))

We need not find 's' to prove that 'a' > 's' becomes too large. We could just find a number 2^r less than 's' but is still large enough so that p^2^r becomes too large for (p-1)!+1. For that, we need to study some properties of 2^r powers of p mod 2^f(p).