I recently started learning number theory, and I've noticed there are many conjectures about the prime numbers that are unproven. Some examples would be whether there are infinite Mersenne, Sophie-Germain, or Fermat primes. There are also problems like the Goldbach conjecture. Yet, primes are defined so simply. Why are they so hard to work with? I am very new to number theory, so I am looking for a simple, intuitive answer.
[Math] Why are conjectures about the primes so hard to prove
elementary-number-theoryprime numberssoft-question
Related Solutions
The standard conjectures imply that there are infinitely many primes of the form $p=n^4+2$. No such prime can be part of a twin pair, since $p-2=n^4$ and $p+2=n^4+4=n^4+4n^2+4-4n^2=(n^2+2)^2-(2n)^2=(n^2+2n+2)(n^2-2n+2)$.
A simpler example is $p=n^2-2$.
An even simpler example is $p=21n+5$, and here we don't need any conjectures --- we know that there are infinitely many such primes $p$. Many more examples can be constructed along the same lines, e.g., $15n+7$, $15n+8$, $77n+9$, $39n+11$, etc., etc.
EDIT: The above was written before OP edited the question to indicate interest only in the five types of prime at the top of the question. So, let's look at those. In all cases, please ignore tiny counterexamples to generally true statements.
Mersenne primes $q=2^p-1$, $p$ prime. Trivially $q$ can't be the smaller of a pair of twin primes, so we are asking about $2^p-3$ and $2^p-1$ both being prime. Apparently this does happen from time to time, so there's no simple reason why it shouldn't happen infinitely often. On the other hand, we don't even know there are infinitely many Mersenne primes, so we're not going to prove it does happen infinitely often. In short: hopeless.
Sophie Germain primes: $p$ such that $p$ and $2p+1$ are both prime. $p-2$ is a multiple of 3, so we are asking whether there are infinitely many $p$ such that $p$, $p+2$, and $2p+1$ are all prime. The standard conjectures (e.g., Schinzel's Hypothesis H) say yes, but no one has a clue as to how to prove this. In short: hopeless.
Fermat primes. Maybe there are some congruences to show $2^{2^{2n}}+3$ can't be prime for sufficiently large $n$. It's worth a look. On the other hand, maybe there are only 5 Fermat primes anyway. There are heuristic arguments suggesting that there are only finitely many.
Regular primes. Another set that hasn't been proved infinite, although the smart money is leaning that way. I can't imagine any relation between the regularity of $p$ and the primality of $p\pm2$. Possibly just ignorance on my part, but I'm going to call this one: hopeless.
Fibonacci primes. The Fibonacci numbers grow exponentially, just like the powers of 2 (only not quite as fast), so the situation here is comparable to that with the Mersenne numbers. Hopeless.
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $n\ge 2$
Proof : Because of $2^4\equiv 1\mod 5$ we can reduce the exponent modulo $4$, which gives $2^1+3=5$ which is divisble by $5$. QED
Hence there is no further prime of the form $2F_n+1$ and therefore no further Fermat-Sophie-Germain-prime.
Best Answer
Here's a brief, vague, and incomplete answer but one that relates to many open problems (and both the Goldbach and Twin Prime conjectures): primality is a multiplicative condition, not an additive one. Understanding how they are distributed in an arithmetic sequence (e.g. in the natural numbers or for use in problems involving prime gaps (roughly what's needed in the above two conjectures)) is attempting to understand how the notion of primality relates to something defined in terms of addition, and there's no obvious way to understand the definition of a prime number in terms of addition.
Edit: I just noticed that this was already mentioned in the comments but may as well leave it here. frogeyedpeas's answer brings up another very good point.