Even integer has at least $K$ representations as the sum of two primes for any $K$ I want to show that for any integer $K$, there exists an even integer such that it can be represented as a sum of two primes in at least $K$ different ways. I'd also like to work on a similar statement for differences between two primes.
I'm aware of the Goldbach conjecture but this seems slightly different since it's asking for the number of different representations for the sum. I've started out with a simple example. For $K=2$, we have $14=7+7=11+3$. I'm not sure this is something I could prove with induction... Any help or hints to get me started?
 A: The prime-counting function $\pi(n)$ is known to lie between $\frac{n}{\log n}$ and $1.25506\frac{n}{\log n}$ for all $n\geq 17$.  [Source: Wikipedia]
This means for sufficiently large $n$, there are at most $1.25506\frac{n}{\log n}$ primes less than $n$, and at most $\frac{2n}{\log 2n}$ primes less than $2n$, that is, there are at least $2\frac{n}{\log 2 + \log n} - 1.25506\frac{n}{\log n}$ primes between $n$ and $2n$.
For any $\epsilon$, we can say that for sufficiently large $n$, $2\frac{n}{\log 2 + \log n}$ is more that $(2-\epsilon)\frac{n}{\log n}$, so there are at least $(2-\epsilon-1.25506)\frac{n}{\log n}$ primes between $n$ and $2n$.
To make things more concrete, choose $\epsilon = 0.00594$, so $2-\epsilon-1.25506=0.74$. Thus, for $n$ large enough, we can be sure there are are at least $0.74\frac{n}{\log n}$ primes between $n$ and $2n$.
There therefore at least $\frac{1}{2}0.74\frac{n}{\log n}\left(0.74\frac{n}{\log n}-1\right)$ ways to sum two prime numbers between $n$ and $2n$, and they form a sum between $2n$ and $4n$.
For sufficiently large $n$, $\frac{1}{2}0.74\frac{n}{\log n}\left(0.74\frac{n}{\log n}-1\right)$ is more than $\frac{1}{2}0.74\frac{n}{\log n}\left(0.72\frac{n}{\log n}\right)$, so there are at least $0.2664\frac{n^2}{(\log n)^2}$ ways to express numbers in the range $2n$ to $4n$ as the sum of two primes.
Since all the sums are even, and there are $n$ even numbers in the range, the pigeonhole principle shows that at least one of the numbers in the range $(n,2n]$ is the sum of two primes in at least $0.2664 \frac{n}{(\log n)^2}-1$ ways.
Since that expression approaches infinity as $n\rightarrow\infty$, this gives the result you want.
