This is a question expecting the answer no. I'm wondering out of curiosity whether there is any positive integer $n$ for which it is known that $2n$ is a sum of two primes, but which is such that no two primes are known that sum to $2n$. A stronger version of the question would be whether there is an integer $n$ that is greater than the largest known prime but for which it is nevertheless known that $2n$ is a sum of two primes.

  • 10
    $\begingroup$ I would not call this "non-constructive" but rather "non-explicit". $\endgroup$ Commented Aug 16, 2023 at 20:24
  • 2
    $\begingroup$ How about this? Write a computer program to generate two large primes. Add them together, save the result, but don't save the primes. $\endgroup$ Commented Aug 17, 2023 at 3:14
  • 3
    $\begingroup$ By "it is known that" I mean that I get to read a proof. (But I realize your suggestion was only semi-serious.) $\endgroup$
    – gowers
    Commented Aug 17, 2023 at 9:28
  • 4
    $\begingroup$ I'm not a fan of these types of questions. A better question would be: is there a known way to determine whether an even number is a sum of two primes other than brute force search? $\endgroup$ Commented Aug 17, 2023 at 14:16
  • 2
    $\begingroup$ Well, almost all even numbers are sums of two primes. Do they qualify? Also, if $2n$ is a sum of two primes, then one also "knows" two such primes, namely the first prime pair (in some explicit ordering) whose sum is $2n$. $\endgroup$
    – GH from MO
    Commented Aug 17, 2023 at 22:15

3 Answers 3


If the even number $2n$ in question has to be chosen in a manner that does not already construct an explicit prime representation $2n = p_1 + p_2$ as a byproduct of the construction of $n$, then I think the answer is "Not without a major advance towards resolving the even Goldbach conjecture". I illustrate this belief with two possible approaches.

  1. (Circle method) It would suffice to get a sufficiently non-trivial lower bound on $$ \sum_{m \leq 2n} \Lambda(m) \Lambda(2n-m)$$ where $\Lambda$ is the von Mangoldt function. This can be written as $$ \int_0^1 S(\alpha)^2 e(-2n\alpha)\ d\alpha$$ where $S(\alpha)$ is the exponential sum $S(\alpha) := \sum_{m \leq 2n} \Lambda(m) e(m\alpha)$ (one can also use a smoother cutoff here if desired). As is well known, the major arc contribution of this integral can be shown to be large and positive (for suitable parameter choices to define ``major arc''), so it would suffice to obtain a sufficiently strong upper bound on the minor arc contribution. Unfortunately, even assuming GRH, the best known bounds on the minor arc contribution (from either Vinogradov bilinear sum methods, or the Plancherel identity) are larger than the expected size of the major arc contribution (and there are multiple obstructions to defeating this, including but not limited to Siegel zeroes and the parity problem). However, one could in principle imagine that for some specific $n$, the minor arc contribution could be estimated numerically, for instance by a Monte Carlo method. But given how difficult it would be to estimate even a single value of $S(\alpha)$ at the required level of accuracy, I doubt that this numeric computation could be any faster than the $O(\log^{O(1)} n)$ (expected) time needed to produce an explicit prime representation $2n = p_1 + p_2$ by selecting $p_1$ randomly and then applying the fastest available primality testing algorithm to both $p_1$ and $2n-p_1$, repeating until a prime representation is found. In principle it could be faster to test some special type of $n$ (e.g., one that is in a specific residue class with respect to small primes), but the speedup here seems very modest.

  2. (Pigeonhole principle) Another possible way to show non-constructively that $2n$ is the sum of two primes is to exhibit some finite set of natural numbers $A$ such that the density of primes on both $A$ and $2n-A$ exceeds $1/2$, as the existence of such a representation would then follow from the pigeonhole principle. But the parity problem defeats any current sieve-theoretic approach towards locating such a set $A$. One could try some sort of random sampling to build such an $A$ instead, but this is basically just a worse version of the aforementioned strategy to find a prime representation $2n = p_1 + p_2$ by testing random decompositions. Again, choosing a special type of $n$ may help a little bit (for instance, if $n$ is divisible by many small primes then the number of representations is slightly elevated) but not enough to make a significant impact, I think.


I think that the answer to the "stronger" question is YES.

If I understand correctly $M_{82 \, 589 \, 933}:=2^{82 \, 589 \, 933}-1$ is the largest known prime number (up to now). The second largest known prime number is $M_{77 \, 232 \, 917}:=2^{77 \, 232 \, 917} − 1$. Thus, $$M_{82 \, 589 \, 933}+M_{77 \, 232 \, 917}$$ is an integer that can be written as the sum of two prime numbers and which is greater than the largest known prime number.

UPDATE: Here is my take on the corrected version of the "stronger" question posed by Professor Gowers (the answer is still a "YES"). According to the groundbreaking papers by Y. Zhang, J. Maynard, etc., there are infinitely many prime numbers which differ by at most $246$. This implies that, for some even integer $c \in [2,246]$, there exist infinitely $k \in \mathbb{N}$ such that

$$p_{k+1}-p_{k} = c$$.

In particular, this implies that we can find a $k \in \mathbb{N}$ such that $p_{k} > M_{82\,589\,933}$. Since

$$p_{k+1}+p_{k} = 2p_{k}+c,$$

it follows that $n=p_{k}+c/2$ is an example of an integer that is greater than the largest known prime number and which satisfies that $2n$ can be written as the sum of two prime numbers.

  • 9
    $\begingroup$ Possibly @gowers should amend his question by saying "... greater than twice the largest known prime..." $\endgroup$ Commented Aug 16, 2023 at 21:17
  • 2
    $\begingroup$ I meant that 2n was a sum of two primes. Corrected now. $\endgroup$
    – gowers
    Commented Aug 16, 2023 at 23:27
  • 1
    $\begingroup$ The second paragraph of your answer can be future proofed by defining $p(t, n)$ to be the $n$th largest prime known at time $t$, with appropriate restrictions on $n$ in terms of $t$, and then inviting the reader to consider $p(t_\text{present}, 2) + p(t_\text{present}, 1)$ at the time $t_\text{present}$ of their reading. $\endgroup$
    – LSpice
    Commented Aug 17, 2023 at 23:00
  • 1
    $\begingroup$ Re, I do not think such questions are appropriate, so I will not answer. But my comment was purely joking, and no malice was meant. $\endgroup$
    – LSpice
    Commented Aug 17, 2023 at 23:44
  • 7
    $\begingroup$ I'm not sure I find this construction a convincing answer to the question. This $n$ does have a "known" representation as the sum of two primes, namely $p_k$ and $p_k+c$; and the procedure used to construct $n$ will also construct these primes as a natural byproduct. I interpret the question as looking for a way to produce an $n$ in a way that one does not automatically also construct a prime representation $2n = p_1 + p_2$ as a byproduct, but for which one can still prove non-constructively that such a representation exists. $\endgroup$
    – Terry Tao
    Commented Aug 18, 2023 at 4:27

We assume that for the largest known prime $p_\text{max}$ it holds that $p_\text{max} < n$ and $2n = p_1 + p_2$. Then it follows that $2p_\text{max} < 2n$.

The Bertrand's theorem states that for each positive integer $m \ge 2$ there is a prime $p$ such that $m < p < 2m$. For $p_\text{max}$ there is a next prime $p_{\text{max}+1}$ and it holds $p_\text{max} < p_{\text{max}+1} < 2 p_\text{max}$.

We set $2n = p_\text{max} + p_{\text{max}+1}$ and it follows that $2n < 2 p_{max+1}$. It implies that a positive integer $n$ between $p_\text{max}$ and $p_{\text{max}+1}$ exists and $2n$ is a sum of two consecutive primes $p_\text{max}$ and $p_{\text{max}+1}$. The prime $p_{\text{max}+1}$ lies between $p_\text{max}$ and $2p_\text{max}$.

  • $\begingroup$ What is your assumption on $p_\text{max}$? Specifically, what are $n$, $p_1$, and $p_2$? Do you mean "there exists an integer $n > p_\text{max}$ such that $2n$ is the sum of two primes"? However interpreted, this seems at best like an existence result (though it seems to be assuming the existence?), not the requested proof that a specific integer satisfies the desired condition; how does it improve on, or add to, the existing answers? $\endgroup$
    – LSpice
    Commented Feb 10 at 22:01
  • $\begingroup$ @LSpice: I understand that a non-constructive proof = a proof of existence is requested. An integer that satisfies the desired condition is specified as $2𝑛=𝑝_{max}+𝑝_{max+1}$. I mean there exists an integer $n$ higher than the largest known prime $𝑝_{max}$ such that $2n$ is the sum of two primes (generally named) $p_{1}$ and $p_{2}$. $\endgroup$ Commented Feb 12 at 22:06
  • $\begingroup$ Re, the question asks for a particular, explicitly given, integer $n$ of which "$\exists\text{$p_1$, $p_2$ prime}\mathpunct.2n = p_1 + p_2$" has been proven, but only non-constructively—a construction of an integer satisfying a property involving non-constructiveness. As far as I can tell, your argument says only that such an $n$ exists, without producing it, which is not what was asked. $\endgroup$
    – LSpice
    Commented Feb 12 at 22:17
  • $\begingroup$ @LSpice: I have produced a particular, explicitly given, integer $n = \frac{p_{max} + p_{max+1}}{2}$. The non-constructiveness lies in an unknown prime $p_{max+1}$. $\endgroup$ Commented Feb 16 at 14:42
  • 1
    $\begingroup$ @LSpice: I see what you mean. Thank you for clarification. $\endgroup$ Commented Feb 16 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.