How often is 2^p-1 square-free

[drew]

Originally Posted by John Renze

It is conjectured that all Mersenne numbers are squarefree.

http://primes.utm.edu/notes/proofs/SquareMerDiv.html

John

only for prime exponents.

9 divides M21.

Drew

[drew]

Originally Posted by drew

only for prime exponents.

9 divides M21.

Drew

Oops, typo. I meant M12.

[druseltal2005]

Originally Posted by drew

Oops, typo. I meant M12.

But M12 is not "Mersenne", since "Mersenne" means: the index is prime.

For any primefactor p of M_n it is:

              n
              ~ ( f + {n,p}')
 n            s
2  - 1  = x* p                        

where

• s is the S-ize (or order) of the multiplicative cyclic subgroup of 2^k (mod p)

• f is the power of p at its F-irst occurence (at n=s)

• n~s is 1 if s divedes n and is 0, if not

• {n,p}' = n~s + n~s² + n~s³ +... = max power of primefactor p in n

So s(3) is 2, and f at (2^2-1 = x*3^1) is 1 and since in your example we have n=12 this is for the primefactor 3:

              12
              ~ ( f + {12,3}')
 12           2
2  - 1  = x* 3                     

              1 ( 1 + 1 ) 
        = x* 3               // 2 divides 12,  and 3 is in 12 to the power of 1

              2
        = x*3      = x* 9                

and thus M12 contains 3 to the power of 2.

But the interesting question is: is there a Mersenne-number (where the index is prime) having any primefactor to a higher power than 1?

[druseltal2005]

Originally Posted by wblipp

I know of three examples of squares dividing p^q-1 for odd p and q less than 100

11^2 | 3^5
7^2 | 67^3
47^2 | 71^23

I also have some examples for larger values of p if that's of any interest. Also a few examples of cubed divisors.

one with exponent 4

2819^3 - 1 = x* 19^4

and forms of the following type have even higher powered primefactors

127^2 - 1 = x* 2^7
255^2 - 1 = x* 2^8
242^2 - 1 = x* 3^5

;-)

[Citrix]

if a==-1 or +1 (mod p^2)

then all a are such that a^p-1 =1 (mod p^2)
I think a must be smaller than p.

Citrix

[Numbers]

Originally Posted by Citrix

Only solution to a^x-b^y=1

(2^1) - (1^2) = 1