# Is any Mersenne number $M_p$ divisible by $p+2$?

More precisely, does there exist a natural number $p$ such that $(2^p-1)/(p+2)$ is also a natural number? It seems to me that this is a really simple problem (with the answer "no"), but I couldn't find anything on the web. There are some facts known about division by $p+1$, but nothing useful for $p+2$.

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Someone should double-check this, but my computer seems to be telling me that $${2^{20735}-1\over20737}{\rm\ and\ }{2^{93525}-1\over93527}$$ are natural numbers. No other examples up to 100001.

It appears that these numbers have been tabulated.

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I think you forgot the $-1$ in the numerator. –  EuYu Aug 30 '12 at 2:58
@EuYu, thanks - fixed now. –  Gerry Myerson Aug 30 '12 at 4:41
I can confirm the computations. Also in case nobody noticed $\frac{2^0-1}{0+2}=0$ is a natural number (to me). –  Marc van Leeuwen Aug 30 '12 at 6:49

If $p$ is even, $p+2$ cannot divide $2^p-1$

Else p is odd.

If $(p+2)\mid(2^p-1),$ $2^p≡1\pmod{p+2}$ but $2^{\phi(p+2)}≡1\pmod{p+2}$

So, $\operatorname{ord}_{p+2}2$ must divide $(\phi(p+2),p)$.

If p+2 is prime, $(\phi(p+2),p)=(p+1,p)=1$.

Else $\phi(p+2)<p$ as in that case $\phi(p+2)<p+1$ and $\phi(p+2)≠p$, if $p$ is odd prime $\implies(\phi(p+2),p)=1$

So if at least one of $p$ and $p+2$ is odd prime, $\operatorname{ord}_{p+2}2\mid1$ which is impossible.

When both of them are composite,

If $p+2=\prod Q_i^{r_i}$ where $r_i$s are positive integers and $Q_i$ are distinct odd primes,

As $(p,p+2)=(p,2)=1$ as $p$ is odd,

so, $(p,Q_i)=1 \implies (p, \phi(p+2))= (p, \prod Q_i^{r_i-1}(Q_i-1))=(p, \prod(Q_i-1))$

If $Q_i-1=2^{s_i}q_i$ where $q_i$ is odd number, $(p, \phi(p+2))=(p, \prod(Q_i-1))$ $=(p, \prod q_i)$ as $p$ is odd.

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Another proof is as follows. First for $p=2$, it follows because $2^p-1=3<4=p+2$. For odd primes $p$, there is a theorem (see Mersenne Prime for proof) that states that every factor of $2^p-1$ is of the form $2kp+1$ for some integer $k\geq0$. Thus we want a $p$ for which $p+2=2kp+1$. There is no such prime because for $p>2$, we have $1<p+2<2p+1$ thus requiring $0<k<1$, a contradiction to the theorem.

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But OP didn't say $p$ was prime. –  Gerry Myerson Aug 30 '12 at 2:44
In fact $p+2=2kp+1 \implies (2k-1)p=1$ –  lab bhattacharjee Aug 30 '12 at 4:58