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I have another question, On the prime divisor of $n=(p^{2}-1)/2$, where $p$ is prime, please tell me your idea. Let $p\neq 3$ be Merssen prime. Is it true $n$ has a prime divisor $r$ such that $r^{2}$ does not divide $n$?

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The condition "for every prime $q$ dividing $n$, $q^2$ also divides $n$" is called being powerful in the literature. So after dividing out the factors of $2$, you'd like to know if $2^{p-1} - 1$ is ever powerful when $2^p - 1$ is prime. Ribenboim has some conjectures related to similar questions in "The New Book Of Prime Number Records", in the section on Wieferich primes. –  Zack Wolske Apr 6 '12 at 18:25
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Well, maybe I'm missing the point, but the first Mersenne prime is $3=2^2-1$, and $\frac{3^2-1}{2} = 4 = 2^2$, so... no, I guess.

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Considering there are only 47 known Mersenne primes, and finding the factors of $2^p-1$ is a difficult task, I'm not sure this question is fully tractable. But we can certainly show that all of the known Mersenne primes satisfy your question. First, take $m=2^p-1$ to be a Mersenne number, and rewrite $n= \frac{m^2 - 1}{2} = 2^{2p-1} - 2^p$. This is powerful when $p=2$, which gives Philip van Reeuwijk's counterexample. For larger $p$, we know $4$ will always divide $n$, so we can ignore powers of $2$ and just consider the odd part $n'= 2^{p-1} - 1$.

We know $p$ is prime, and odd since it is not $2$, so we are looking at $n' = 4^k - 1$, with $k=\frac{p-1}{2}$. Clearly $3$ divides $n'$, and $9$ divides iff $3$ divides $k$. So we require that $p \equiv 1 \mod6$. This is not enough; plenty of known Mersenne primes have this property.

Since $3$ divides $k$, we can write $n'=64^{k'}-1$. Then $7$ divides $n'$, and $7^2$ divides $n'$ iff $7$ divides $k'$. So now we require that $p \equiv 1 \mod42$. Sadly, again this is not enough.

One step further, we see that when $7$ divides $k'$, $43$ divides $64^{k'}-1$, and $43^2$ divides iff $43$ divides $k'$. Now we're happy (for the time being), because no known Mersenne prime has $p\equiv 1 \mod 1806 = 43*7*6$. But it seems there's no reason they can't have this property, so you may have to continue your search once such a Mersenne prime is found. Expect one by the 504th instance: $504 = \phi(1806)$.

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I have checked a few cases, and if $m = 2^p-1$, it actually seems the case that $p$ is a factor of $n = \frac{m^2-1}{2}$ and $p^2$ isn't. This even seems to work when $m$ isn't a prime but just the $p$-th Mersenne number with $p$ prime. I can't prove it, though... –  Philip van Reeuwijk Apr 6 '12 at 19:32
    
If you could that would be something interesting. Showing that $p$ divides is an application of little Fermat. Conjecture $W'_2$ on $p. 343$ of Ribenboim's "New Book of Prime Numbers" is "There exist only finitely many primes with $2^{p-1} \equiv 1 \mod p^2$", and he goes on to show that this would imply solutions to other well-known open problems. –  Zack Wolske Apr 6 '12 at 20:35
    
Yes, the little Fermat part I had figured out already; the part about $p^2$ not being a factor is beyond me, I fear. Ribenboim's book looks interesting, I'll check it out. Thanks for the tip! –  Philip van Reeuwijk Apr 6 '12 at 20:52
    
Just to clarify, such primes are called Wieferich, and only two are known: 1093, and 3511. –  Zack Wolske Apr 6 '12 at 21:53
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