# Are there any intuitive reasons for Goldbach conjecture to be true?

One thing puzzled me is that, despite its simple form, I have not seen any intuitive reasons for Goldbach conjecture to be true.

Typical heuristic reason is based on probability arguments. Such arguments basically say that, since there are so many primes out there, so very likely, you will find a Goldbach combination for any even number (2n = p1 + p2).

Circle methods and sieve methods essentially follow above argument.

The issue with this argument is that, it can not explain why Goldbach conjecture hold true for small numbers. For example, why is it true for all even numbers less than 100 or 1000.

Are there any intuitive reasons for Goldbach conjecture to be true ?

• It isa "soft" mathematical question - what do you mean by "intuitive" - there are many statements which supports the conjecture, computer powered methods still did not overrule it - however a single number printed in one text line can be a counterexample. Maybe the best would be to follow the consequences coming from Goldbach conjeture being true or false. – z100 Nov 9 '15 at 23:18
• Well, yes, it could have been false for small values. But it just happens to be true. Would it be false for small values the conjecture might be every even $n$ greater than $124$ is the sum of two primes. – quid Nov 10 '15 at 0:45
• Look at it this way: There are only five numbers below $10^{10}$ which are not the sum of a prime and a perfect power. – Lucian Nov 10 '15 at 7:15

It is intuitive to me, for instance, for one specific reason. Think of even numbers (under a certain $N$) as spaces, like, rooms. Primes which are not larger than $N - 3$ are so populous that their pairwise sums hardly find enough room to fit, and sometimes several of such sums have to get in the same room. To see this, let $k$ be the number of odd primes up to $N - 3$, then there are ${k \choose 2} = \frac {k^2 - k} {2}$ pairwise sums of primes which do not exceeds $N$, (note that some, and actually most, of them will be repeated) which is around $\frac {N^2} {2 \log ^2 N}$, which is $\gg N$. Hope this helps.