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To give a vague answer, I think these questions are difficult because they mix multiplicative conditions (being prime) and additive conditions (as in the twin prime case).

Basically all results about primes that I can think of come down to unique factorization of the integers. For example, the zeta function is given as

$$\zeta(s) = \sum_n n^{-s} = \prod_p (1 - p^{-s})^{-1}.$$

The right hand side is why the zeta function tells you about prime numbers, but the left hand side is what typically helps you prove theorems. For example, Riemann noticed that the left side looked like something similar to what Poisson summation is good for, and hence proved analytic continuation and the functional equation.

On a simpler level, one nice proof that there are infinitely many primes is to observe that $\sum_n 1/n$ diverges, by elementary calculus, and therefore the right hand side diverges for $s = 1$ as well.

Gerhard Paseman suggested looking at arithmetic progressions, and I think this is an extremely instructive example. Looking at the sum of $n^{-s}$ restricted to an arithmetic progression, you don't have any equation like the above. Conversely, if you take a product over only the primes $p$ in some arithmetic progression, you don't get anything nice like the left side. However, if you let $\chi$ be a Dirichlet character, e.g., a homomorphism from $(\mathbb{Z}/N)^{\times}$ to $\mathbb{C}$, then you get the Dirichlet $L$-function

$$L(s, \chi) = \sum_n \chi(n) n^{-s} = \prod_p (1 - \chi(p) p^{-s})^{-1}.$$

In some way this is forcing a round peg into a square hole: the arithmetic progression condition couldn't be handled directly. But it can be written as a linear combination of Dirichlet characters, and once you force everything to be multiplicative, the machinery (Poisson summation, etc.) all works.

So in other words, IMHO, the question isn't "why is the twin prime conjecture difficult", but "why can we prove anything about the primes at all?" Our toolbox is, in my experience, still pretty limited.

1

To give a vague answer, I think these questions are difficult because they mix multiplicative conditions (being prime) and additive conditions (as in the twin prime case).

Basically all results about primes that I can think of come down to unique factorization of the integers. For example, the zeta function is given as

$$\zeta(s) = \sum_n n^{-s} = \prod_p (1 - p^{-s})^{-1}.$$

The right hand side is why the zeta function tells you about prime numbers, but the left hand side is what typically helps you prove theorems. For example, Riemann noticed that the left side looked like something similar to what Poisson summation is good for, and hence proved analytic continuation and the functional equation.

On a simpler level, one nice proof that there are infinitely many primes is to observe that $\sum_n 1/n$ diverges, by elementary calculus, and therefore the right hand side diverges for $s = 1$ as well.

Gerhard Paseman suggested looking at arithmetic progressions, and I think this is an extremely instructive example. Looking at the sum of $n^{-s}$ restricted to an arithmetic progression, you don't have any equation like the above. However, if you let $\chi$ be a Dirichlet character, e.g., a homomorphism from $(\mathbb{Z}/N)^{\times}$ to $\mathbb{C}$, then you get the Dirichlet $L$-function

$$L(s, \chi) = \sum_n \chi(n) n^{-s} = \prod_p (1 - \chi(p) p^{-s})^{-1}.$$

In some way this is forcing a round peg into a square hole: the arithmetic progression condition couldn't be handled directly. But it can be written as a linear combination of Dirichlet characters, and once you force everything to be multiplicative, the machinery (Poisson summation, etc.) all works.

So in other words, IMHO, the question isn't "why is the twin prime conjecture difficult", but "why can we prove anything about the primes at all?" Our toolbox is, in my experience, still pretty limited.