Suppose $a,n \in \Bbb N $. How do you prove that if $a^n-1$ is prime, then $a=2$ and $n$ is prime .
Is the converse true I.E : (if $p$ is prime then $2^p-1$ is prime)?
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2Don't forget $a = 3$, $n = 1$!. – Ryan Reich Mar 12 '13 at 18:45
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If $n=1,a-1$ will be prime $=p$(say) if $a=p+1$ where $p$ is any prime.
For $n>1,$
As here $(a-1)\mid(a^n-1)$
$a-1$ must be $<2\implies a=2$
Now, if $n=rs$ where $r,s$ are integers$>1$,
$(a^r-1)\mid (a^{rs}-1)\implies a^n-1$ can not be prime
So, the necessary condition for $a^n-1$ be prime are $a=2$ and $n$ is prime.
The converse is not true as for example $2^p-1$ is not prime for $p=11$
This using $3$rd theorem of this, if prime $q\mid(2^p-1), 2p\mid(q-1)$ i.e., $q=2\cdot k\cdot p+1$ where integer $k\ge0$
For $p=11, q=22k+1$ and we can verify $2^{11}-1=23\cdot89$
Reference: List of known Mersenne primes
lab bhattacharjee
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Notice that, when $a$ is odd, $a^n-1$ is even, thus =$2$, i.e. $a=3, n=1$. Otherwise this proof is perfect. – awllower Mar 13 '13 at 13:05
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@awllower, for $n=1,$ if $a=p+1$ where $p$ is any prime, $a=p$(prime). I think, I should include the assumption that integer $n>1$ – lab bhattacharjee Mar 13 '13 at 15:07
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It is always a good thing to improve upon the completeness, at least from my point of view. Thanks for the answer BTW. – awllower Mar 13 '13 at 15:21
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You will find that $23$ divides $2^{11}-1$.
For much more, search for Mersenne prime. From the computations that have been done, $2^p-1$ is composite for "most" primes $p$. The finding of a Mersenne prime is a fairly rare event. But it just happened again recently.
André Nicolas
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