second time puzzled.

Last post we inched closer to the proof that

For prime p ,

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

is only possible for p <=5. We will consider four cases of primes p = 8n±1 and p = 8n+3 and p = 8n+5 to study how 2^r powers of p behave `mod` 2^f(p).

For p = 8n±1 we extract all powers of 2 from n and write

p = 2^x * (2*m+1) ± 1

p^2 = ( 2^(x+1) + 1 ) `mod` 2^(x+2)

p^2^r = (2^(x+r) + 1) `mod` 2^(x+r+1) where r > 0.

Now substitute y = x+r+1 in the above

p^2^(y-x-1) = (2^(y-1) + 1) `mod` 2^y

holds provided y > x+1

For p = 8*n + 5 and p = 8*n+3,

p^2 = (2^3 + 1 ) `mod` 2^4
p^2^r = ( 2^(2+r) + 1 ) `mod` 2^(3+r)

Now substitute y = 3+r,

 p^2^(y-3) = ( 2^(y-1) + 1 ) `mod` 2^y 

and so

 p^2^(y-3) = ( 2^(y-1) + 1 ) `mod` 2^y 

Lets gets back to the proof to where we had left it that is to get a lower bound on 's' such that

 p^s = 1 mod 2^f(p) 

If we find a number 2^r such that p^2^r ≠ 1 mod 2^f(p) it would mean s > 2^r.

From what we saw above, if 2^y is some power of 2 that divides (p-1)!,
for p = 2^x*(2*m+1)± 1

 p^2^(y-x-1) = (2^(y-1) + 1) `mod` 2^y 

and integer y > x+1, x >= 3

 s > 2^(y-x-1) 

and since `a' is a multiple of s,

 a > 2^(y-x-1)

and so

p^2^(y-3) = (2^(y-1) + 1 ) `mod` 2^y 

for p of the form 8n+3 or 8*n+5

s > 2^(y-3)

and since `a' is a multiple of s,

a > 2^(y-3)

From my previous post , y could be chosen

 y = (p - 1 - 2^n)/2 + (2^n -1 )

where 2^n is greatest 2 power less than or equal to (p-1)

for p = 2^x*(2*m+1) ± 1,

 a >= s > 2^( (p - 1 - 2^n)/2 + (2^n -1 ) - x -1 ) = 2^(2^(n-1) + (p-1)/2 - 2 - x ) 

and for p = 8n+3,8*n+5

 a >= s> 2^( (p - 1 - 2^n)/2 + (2^n -1 ) - 3 ) = 2^( 2^(n-1) + (p-1)/2 - 4 ) 

for p = 11 = 8*1 + 3 n = 3 ,

  a > 2 ^( 1 + 8 -1 -3) = 16 

for p = 13 = 8*1 + 5 , n = 3

 a > 2 ^( 2 + 8 -1 -3) = 32 

for p = 17 w = 4 , x = 4

a > 2^( 16 - 1 - 4 -1 ) > 2^10

In fact for all primes p >= 17 and p < 32 , a > 2^10 and likewise for primes p between 33 and p < 64 ,

 a > 2^(32 - 2 - 5 )= 2^25 

and in general for primes

 p >= 2^x+1 and p >2^(x+1)

a > 2^(2^x-x-2)

So all we now need show is

 2^x > 2*x+3

for x = 4 it is obviously true.
if it true for some m.

 2^m > 2*m+3

we show it is true for m+1, as

 2^(m+1) = 2 * (2^m) > 4*m+6 = 2*(m+1) + 3 + (2*m+1) > 2*(m+1) + 3.

2^x > 2*x+3 is true for x >=4

then

2^x - x - 2 > x + 1

raising both sides to 2 power

2^(2^x-x-2) > 2^(x+1)

Now since p lies betwen 2^x + 1 and 2^(x+1)

a > 2^(2^x-x-2) > 2^(x+1) > p

which is too large for p^a = (p-1)!+1.

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