Problems & Puzzles: Conjectures Conjecture 12. Are Mersenne and Fermat numbers square free ? A "squarefree number" is a composite number which has no repeated factor. Well, nobody knows if the Mersenne numbers (2^{p} 1, p=prime) or the Fermat numbers are squarefree or not… A close problem related with this one is the following : From the Fermat (little) theorem we know that if p is prime then p divides 2^{p} 2. But how many primes p<10^{6} are such that p^{2 }divides 2^{p} 2. The answer is p= 1093 and p=3511, only. Now (following a challenge of Selfridge) would you like to compute fifty more primes p such that p^{2 }divides 2^{p} 2 ?… (Ref. 2, pp. 89) Solution Mr. Jacques Basaldúa, from Spain, sent (16/12/2001) the following message:
*** But, this proof has a flaw, as Mr. Pavlos N saw (message 4609):  In primenumbers@y..., Pavlos N <pavlos199@y...> wrote: > Can anyone comment or find a (possible) flaw in : > http://www.dybot.com/numbers/sqfree.htm Yes, there is a flaw. I can't pinpoint the exact error in reasoning (basically, I don't want to spend the time to find it :), but the equation which is "proven" in section 4.2 is not true. There is an easily computable counterexample. The author states that: ß(p^n) = p^(n1)·ß(p) (for any prime p) This is not true if p = 1093 & n = 2: ß(p^n) == 364 p^(n1)·ß(p) == 1093*364 I believe that in general, the postulate in section 4.2 is not true for any Wieferich prime. Somebody apparently tried to prove the Mersennes squarefree using this technique in '96 and came up against this same problem: http://www2.netdoor.com/~acurry/mersenne/archive2/0037.html ***





