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?
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?
It is false. Counterexample: $73$.
$$73 - 2 = 71\;\text{is prime}$$
$$73 = 8^2 + 3^2$$
$$73^2 = 55^2 + 48^2$$
$$2^{73} - 1 = 439×2298041×9361973132609$$
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.
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.
$2^p-1$ can be only a prime number if $p$ is prime, but not for every prime number $p$ is $2^p-1$ a prime number.
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.