# Number of ways to write $2n$ as sum of two primes is unbounded

For all $$k>0$$, we are to show that
there exists an even number $$2n > 0$$ that can be written as sum of two primes in at least $$k$$ ways.
($$8=3+5=5+3$$ counts as one way)

I tried to use pigeonhole principle to "fill" odd numbers in the lower half of $$2n$$ with primes, i.e. to show $$\pi(2n)-[\frac{n}{2}]$$ can be arbitrarily large, but it turns out to diverge to negative infinity instead.
Of course, requiring the entire half to be primes is an overkill, but I don't know how to rephrase the problem into a manageable form.
It probably has something to do with Prime Number Theorem, but I am not sure.

Thank you for any help.

• The Green-Tao theorem states that the sequence of prime numbers contains arbitrarily long arithmetic progressions. Consequently, for any integer $k>0$ there is an arithmetic sequence of odd prime numbers at least $2k$ in length. The sums of the first and last, the second and next to last, etc., will all be the same, and will be an even number. Thus there will be $k$ ways to express that sum. – Keith Backman Nov 25 '20 at 16:16
• @KeithBackman While that’s true and correct, it is a bit like swatting a fly with a high-powered laser. The pigeonhole argument goes through with substantially less input than Green-Tao, and in fact doesn’t even require PNT. It should be enough to use the divergence of the series of prime reciprocals, which dates back to Euler. – Erick Wong Nov 26 '20 at 5:04

You're on the right track with the pigeonhole principle, but your counting is wrong; you want to get a 'quadratic-like' term to show unboundedness. Try this approach instead: consider how many unordered pairs of primes $$\leq n$$ there are; call this $$f(n)$$. Now, note that there are only $$n$$ even numbers less than $$2n$$. If each of these could be written as the sum of primes in $$\leq C$$ different ways for some constant $$C$$, what would that say about $$f(n)$$? Can you find a contradiction from here?
There are $$\pi(n)-1 \approx \pi(n)$$ odd primes less than $$n$$. Summing them two at a time, we obtain $$\frac{(\pi(n))!}{(\pi(n)-2)!2!}=\frac{(\pi(n))\cdot (\pi(n)-1)}{2}$$ even sums, all of which are smaller than $$2n$$. In addition, there are $$\pi(n)$$ occurrences of sums of the form $$p_i+p_i$$ not generated by the pairwise summing, and including those with the previous result, we obtain $$\frac{(\pi(n))(\pi(n)+1)}{2}\approx \frac{(\pi(n))^2}{2}$$ even sums to consider. There are $$n$$ even numbers smaller than $$2n$$
If for a chosen $$n$$ and any $$k$$, the number of sums exceeds $$kn$$, then by the pigeon hole principle, at least one even number $$\le 2n$$ is expressible as a sum of two odd prime numbers in at least $$k$$ different ways. This occurs with certainty when $$\frac{(\pi(n))^2}{2}>kn$$ or $$(\pi(n))^2>2kn$$
According to PNT, $$\pi(n) \approx \frac{n}{\ln(n)}$$. Substituting, the necessary condition becomes $$\frac{n}{\ln^2(n)}>2k$$
Since $$\frac{n}{\ln^2(n)}$$ grows without bound, it is always possible to choose an $$n$$ large enough to ensure that some even number smaller than $$2n$$ can be expressed as the sum of two odd primes in at least $$k$$ different ways.