20200905, 23:16  #1 
"Viliam FurÃk"
Jul 2018
Martin, Slovakia
706_{10} Posts 
All Mersenne numbers Mp are square free (maybe)
Suppose there is a factor q of Mp, q = 2*k*p+1, which is a square of x (x also has to be a factor of Mp):
2*k*p+1 = x^{2} 2*k*p = x^{2}  1 2*k*p = (x  1)(x + 1) Let y = x  1 2*k*p = y(y + 2) 2*k*p = y^{2} + 2*y Let x = 2*k_{1}*p + 1, then y = 2*k_{1}*p. 2*k*p = (2*k_{1}*p)^{2} + 2*(2*k_{1}*p) 2*p*(k/2*k_{1}*p) = 2*k_{1}*p + 2 p*(k/2*k_{1}*p) = k_{1}*p + 1 p*(k/2*k_{1}*p)  k_{1}*p = 1 p*(k/2*k_{1}*p  k_{1}) = 1 p*(k/y  k_{1}) = 1 As k must be divisible by y ([2*k*p = y(y + 2)] > Neither 2 nor p can be divisible by y, so k must be), the numbers in the last row are whole numbers. It can be rewritten as p*(a  b) = 1, with p being prime, and a,b being whole numbers But as this can't happen, no Mp can contain squares in its factorization.  Because it's 1:15 here, as I am writing, I can not be sure with my maths. Is there a mistake in my calculations? If yes, where? 
20200906, 02:34  #2  
Jun 2003
2^{3}×11×59 Posts 
Quote:
y does not have to divide k. 2*k*p = y(y + 2) = 2*k1*p*(y+2) Therefore, k = k1*(y+2). i.e. Only k1 needs to divide k p*(k/y  k1) = 1 ==> For this we need that p*k/y be an integer. Therefore p*k/(2*k1*p)= k/(2*k1) be an integer. Thus we can also say that 2*k1 must divide k. Thus your assertion that a=k/y must be an integer is wrong. 

20200906, 13:05  #3 
Feb 2017
Nowhere
7·733 Posts 
If p is prime, and q = 2*k*p + 1 is a prime factor of M_{p}, and q^{2} also divides M_{p}, then (since M_{p} divides M_{q1}), q^{2} divides M_{q1}, i.e. q is a Wieferich prime (to the base 2). The only such primes known are 1093 and 3511. In both cases, the multiplicative order of Mod(2,q) is composite.
Although the search for such primes has been pushed to 10^{17} without finding any more, it is AFAIK not known that there are infinitely many nonWieferich primes to the base 2. This article may be of interest in this regard. 
20200908, 14:48  #4 
Sep 2002
Database er0rr
5×787 Posts 
See my latest test: here and here.
If M_n was a square then for all prime N> some bound kronecker(M_n,M_N) [M_n not square itself] would always be 1, which to me seems counterintuitive to what seems to be "random" with a probability of 1/2 for M_N. Edit there is simpler argument: 2^n1 =s^2 then 2*(2^(n1)1)==(s1)*(s+1), but 2^(n1)1 is odd. So the number of 2's dividing R.H.S. is >2. When I saw "squareless" I did not realize the OP meant squarefree. Last fiddled with by paulunderwood on 20200908 at 15:39 
20200909, 05:18  #5 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
1. Yep, we call these numbers "square free", not "squareless". I modified the thread title, for future generation's searches.
2. There are many "proofs" of the fact that all mersenne numbers with prime exponent are square free (I had one myself, many years ago, on this forum, too ). They all fail in two ways, either they don't use the fact that the exponent is prime (and this is easy to combat, because if the exponent is not prime, then there are easy counterexamples, like 2^61, or 2^211, just to pick the first from the even/odd classes), or either they come from people who never heard in their life about Wieferich primes before creating the proof (my case). If you search the forum for "wieferich" (look for "laurv" as a poster") or "wallsunsun" you can find many "interesting" discussions (no, not this post , this contains wieferich, but the context was different, haha). I remember that there was a guy here (I think he is still a member, but he got a one week ban in 2017 or so, for insulting other forum members, after which he didn't come back, even if the ban expired) who tried to prove there are no WSS primes, based on the fact that if the FLT was false for some prime p, then p is a WSS prime. So, his logic went like "well, FLT is true, then it can't exist any wss prime". He also forgot that if FLT was false for some prime, than that prime was a Wieferich prime, and by his logic, he just proved there are no Wf primes either (two of them are known!). Moreover, he also proved there are no primes (of any kind!) by the same logic (the WSS part was not needed in his "proof": if FLT is false for some exponent, then it is also false for some prime factor of that exponent. Therefore, because Wiles proved FLT to be true, it means there are no primes). Last fiddled with by LaurV on 20200909 at 05:35 
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