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Let $p$ be an odd prime such that $$p \equiv 1 \pmod{4}$$ and $p$ and $p-2$ form a twin prime pair.

My question: Is it true that $2^{p}-1$ is a prime number?

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  • $\begingroup$ If there exists an integer $r > 1$ such that $r^n - 1$ is a prime number for some integer $n > 1$, then $r = 2$. Otherwise $r-1\,|\,r^n-1$ and thus $r^n-1$ is composite. $\endgroup$ – Ben Frankel Oct 13 '14 at 14:46
  • $\begingroup$ @BenFrankel: I have corrected the question. $\endgroup$ – DER Oct 13 '14 at 14:48
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    $\begingroup$ One of your conditions is unnecessary! If $p$ is a sum of two squares then it is automatically the hypotenuse of a right angled triangle... $\endgroup$ – fretty Oct 13 '14 at 14:54
  • $\begingroup$ You can delete the hypotenuse condition and also replace the first condition by $p \equiv 1 \bmod 4$ (since by Fermat we have that $p$ is a sum of two squares if and only if $p=2$ or $p\equiv 1 \bmod 4$, and clearly $2^2-1 = 3$ is prime). $\endgroup$ – fretty Oct 13 '14 at 14:56
  • $\begingroup$ @fretty: But $p>2$. $\endgroup$ – DER Oct 13 '14 at 15:03
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You are saying "if so-and-so is true, does it follow that $2^p-1$ is a Mersenne prime? "

Of the numbers of the form $2^p-1$, only very few are primes. Most exponents p up to 50 million have been examined and it was found in most cases that $2^p-1$ is not prime; there are less than 50 known exceptions. That information is easy to find.

You could just use this information, try out all twin primes (p-2, p) where p = 1 (modulo 4) and see quite quickly that $2^p-1$ is actually very rarely a prime. The first four values for p are 5, 61, 73, 109, and only the first two give Mersenne primes.

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It is false. Counterexample: $73$.

$$73 - 2 = 71\;\text{is prime}$$

$$73 = 8^2 + 3^2$$

$$73^2 = 55^2 + 48^2$$

$$2^{73} - 1 = 439×2298041×9361973132609$$

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Note that $r^{p}-1=(r-1)(r^{p-1}+...+r+1)$. This can only be prime if $r=2$.

The question you are asking is if those conditions imply that $2^p-1$ is a mersene prime.

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  • $\begingroup$ @ N.S.: I have corrected the question. $\endgroup$ – DER Oct 13 '14 at 14:49
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$2^p-1$ can be only a prime number if $p$ is prime, but not for every prime number $p$ is $2^p-1$ a prime number.

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Is it always true? No. (see above answers)

Is it possible? Yes. And perhaps infinitely. Here are the known examples:

$3, 5, 7, 13, 17, 19, 31, 61, 107, 521, 1279, 4423, 110503, 132049, 20996011, 24036583, 74207281$

And those $p=1\pmod 4$ and $p-2$ is prime:

$5, 13, 61, 132049, 74207281$

should account for (asymptotically) $1/4$ of the total twin prime Mersenne exponents since $p+2$ and or $p-2$ could be prime, and $p=1$ or $3 \pmod 4$, leading to 4 different possibilities.

It is also reasonable to conjecture there are infinitely many such primes (both lists):

The sequence of primes $p_n$ (the $n$-th Mersenne prime), appears to be exponential (see this page).

In particular, the geometric mean between two mersenne prime exponents appears to be close to $1/e^γ=1.47576$, so the $n$-th Mersenne prime is modeled by $p_n≈e^{-γn}$ or $1.47576^n$, and therefore reasonable to assume that there exist infinitely many Mersenne primes. The probability that either $p_n±2$ is prime, is close to $3/γn$ or $5.19736/n$ by the Prime Number Theorem. Note that I included constant $3$ in the probability, since, for all primes $p_n$ except $3$, either one of $p_n+2$ or $p_n-2$ is divisible by $3$ and cannot be prime, while the other is not divisible by $2$ or $3$, so the probability is $3$ times as likely to be prime than the ordinary integer.

So really, the probability $p_n$ is a twin primes is $1/n$ times a constant! Going back to the assumption there are infinitely many Mersenne primes, we should expect,

$$C\sum_{k=1}^n \left(\frac{1}{k}\right)$$

Mersenne prime exponents which are also twin primes, among the first $n$ Mersenne prime exponents. Since the sum of $1/k$ is well known to diverge, then the conclusion is that there are infinitely many twin primes $p$ such that $2^p-1$ is prime.

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