second time puzzled.

The problem were trying to prove is if (p-1)!+1 is a power of p
for p > 7. We have already seen the proof for p=7 here

In the spirit of the proof for p = 7 , we'll show that a is greater
than p, but first we need to answer this question:

If 2^f(p) is the greatest 2 power that divides (p-1)! then

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

but what is the value of f(p) ?

If 2^n is the greatest power of 2 less than or equal to (p-1) then

f(p) = floor(( p -1 )/2) + floor(( p -1 )/4) + floor(( p -1 )/8) + ... floor((p-1)/2^n)

We will satisfy ourselves with a lower bound on f(p)

since 2^(n+1) > p-1 >= 2^n

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

=> floor((p-1) / 2^n) = 1

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

But wait, we are overlooking many 2 factors between 2^n and p-1.

which is atleast (p - 1 - 2^n)/2

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

Thats enough for the proof which is to follow in the next post.