Why is the twin prime conjecture hard?

Question

If $\pi_2(x)$ is the number of twin primes of magnitude less than or equal to $x$. We want to prove that $$\lim_{x\,\to\,\infty}\pi_2(x)=\infty$$ which should be easier than finding and proving an asymptotic formula like $x/\log(x)$ for $\pi(x)$. How is it that modern mathematics cannot prove even the awful estimate $$\pi_2(x)\ge \log\log\,...\log x\quad\quad x\text{ big enough}$$ with $100$ or $1000$ nested $\log$s? Any function with unbounded growth (no matter how slow) does the trick. It seems so weird that no argument, sieve-theoretic nor analytic or algebraic, can prove such a "simple result".

I am asking for a concrete obstacle in a would-be proof of this type.

Thanks!

Answer 1

Most of the ways of proving the infinitude of primes rely on their being a basis of the set of natural numbers under multiplication — that is to say, that they generate all the naturals. For instance, the Euler product for the Zeta function takes this form: it's essentially an analytic statement of the fact that each natural number has a unique prime factorization and each product of (possibly repeated) primes corresponds to a single natural number. There's also an information-theoretic proof that goes through this notion of basis: if there were a finite number of primes $p_1, p_2, \ldots, p_m$, then we could write any number $n$ as $p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ and represent $n$ via the list of exponents $\langle e_1, e_2, \ldots, e_m\rangle$. But each of these is at most $\log_2(n)$ and can therefore be represented in $\log_2\log_2(n)$ bits, so that $n$ itself could be represented in $m\lg\lg n$ bits, contradicting the theorem that there are numbers $n$ which require at least $C\lg n$ bits to describe.

But none of this machinery reflects over to the question of twin primes, and worse, twin primes require (gasp) addition, something that plays notoriously poorly with factorization — the abc conjecture being maybe the canonical example of it. The factorizations of $n$ and $n+1$ seem largely independent (aside from, of course, not containing any of the same primes) and likewise the factorizations of $n$ and $n+2$, but as of right now none of the machinery we have really serves to prove this, and so this independence is entirely conjectural.

Answer 2

Here is a funny thought of my own about the counting of twin primes.

An awkward counting formula involving the floor function can be given in the following form, $$\pi_2(x)=\sum\limits_{\substack{1< n\leq x\\n\equiv1\pmod{2}}}\left\lfloor\frac{\varphi(n(n+2))}{n^2-1}\right\rfloor$$ Where $\varphi$ is the Euler totient function.

Now note that we can not approximate $\lfloor x\rfloor$ by $x-1$ because $\left\lfloor\frac{\varphi(n(n+2))}{n^2-1}\right\rfloor$ is either $0$ when $n,n+2$ are not both primes or $\left\lfloor\frac{\varphi(n(n+2))}{n^2-1}\right\rfloor$ is $1$ when both $n,n+2$ are primes. Then this is a big obstacle to find approximate or asymptotic formula for $\pi_2(x)$ which grows to infinity as $x$ gets bigger.

Answer 3

This is a long comment.

Primes are reluctant to yield their secrets. Their distribution is maddeningly irregular. Many Q's on whether there are infinitely many primes of a given type are unanswered, e.g. : Fermat primes, Mersenne primes, Sophie Germain primes (in the early 1800's Sophie Germain proved that Fermat's Last Theorem holds for a prime exponent $p$ when $2p+1$ is also prime.)

The Prime Number Theorem is extraordinarily hard to prove. Even Gauss, who noted it as an idea in his diary, did not prove it.

If $p_n$ is the $n$th prime, it was only in 2013 that it was shown that $p_{n+1}-p_n$ does not tend to $\infty$ as $n\to \infty$. It was shown there are infinitely many $n$ for which $p_{n+1}-p_n<70,000,000.$ The $70,000,000$ has since been improved to about $313$.

It has been shown that the sums of the reciprocals of prime pairs is finite, but this does not tell us whether there are infinitely many of them.