# A hypothesis about the conjecture: “Every even number is the difference of two primes”

If the conjecture "Every even number is the difference of two primes" holds then we conclude the following hypothesis:

Hypothesis. For every distinct non-zero integers $a, b$, at least one of the numbers $a, b$ and $a-b$ can be expressed as the difference of two primes.

Question. Is the converse true (does the hypothesis imply the conjecture)?

• hint : every even number can be written as $a-b$ where $a+2,b+2$ are odd and not prime. – reuns Sep 7 '16 at 20:25

Yes. Take any even $n$; if it were the case that for all odd $b$, at least one of $b+2$ and $b+n+2$ were prime, the primes would have nonzero density.° So take a $b$ for which it is not the case, and the hypothesis implies $n$ is a difference of two primes.
° Partition the odd numbers in pairs with difference $n$: $\{1,n+1\}\{3,n+3\},\ldots,\{n-1,2n-1\};\; \{2n+1,3n+1\},\ldots,\{3n-1,4n-1\};\;\ldots$. There are $kn/2$ pairs below $2kn$ ($k\in\mathbb N$), so there are at least $(x-2n)/4$ pairs below $x$. Each pair has at least one prime, so $\pi(x)\geq x/8-n/4$.
• I am not sure I get it. Let $\mathcal{P}$ be the set of primes and $2m$ some even integer. If $\mathcal{P}$ does not intersect $\mathcal{P}+2m$, where is the contradiction? – Jack D'Aurizio Sep 7 '16 at 20:31
• @JackD'Aurizio I get a contradication from assuming that $p$ or $p+n$ is prime for every odd $p$. There must be a $p$ for which this is not true, and choosing $a=p+n-2$, $b=p-2$ (which are not differences of primes by the choice of $p$) we get that $n\in\mathcal P-\mathcal P$. – rabota Sep 7 '16 at 20:39
• @JackD'Aurizio see my comment, it is clear, and barto used the 0 density of $\mathcal{P}$ for proving my hint – reuns Sep 7 '16 at 20:39
• Apologies, I thought that was intended as a proof of the fact than $2\mathbb{N}\subset (\mathcal{P}-\mathcal{P})$. – Jack D'Aurizio Sep 7 '16 at 20:41