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

Question

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 ?

Answer 1

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.

Answer 2

One way can be to look at the number of satisfying prime-number pairs present for each even whole number greater than $2$. For example, there are $4$ pairs of prime-numbers that satisfy the conjecture for the number $42$ (say), namely $(5,37), (11,31),(13,29)$ and $(19,23)$ but as we move forward on the number line, we have $10$ pairs for the number $162$.

Now, if we graph the number of prime-number pairs that satisfies the conjecture for every even whole number $N$ $(10000>N>2)$, we get an increasing plot with some upper and lower bounds:

Thus, looking at the behaviours of smaller numbers, it makes intuitive sense to me that even for bigger numbers the plot will continue to be increasing and with some lower boundary greater than zero.

Answer 3

An intuitive reason why Goldbach's conjecture holds true for small numbers is as follows:

• if $ 2n = 0 \bmod 6 $ then $ 2n = n_a + n_b $ with $ n_a = 1 \bmod 6 $ and $ n_b = -1 \bmod 6 $ it is possible to verify that for each number $ 2n <966 $ the number total number of primes for $ n_a $ is greater than the total number of composites for $ n_b $ or that the total number of primes for $ n_b $ is greater than the total number of composites for $ n_a $. For this reason there is always at least one solution.

example

$2n=96$ for $n_a$ prime $n_a={7,13,19,31,37,43,61,67,73,79}$ total number of primes for $ n_a $ is $10$ is greater than the total number of composites for $ n_b $ which is $3$ for $ n_b ={35,65,77}$ in fact we have $2n=96=7+89=13+83=37+59=43+53=67+29=73+23=79+17$ , then number of pairs is $7=10-3$.

And also in this case for $n_b$ prime $n_b={5,11,17,23,29,41,47,53,59,71,83,89}$ total number of primes for $ n_b $ is $12$ is greater than the total number of composites for $ n_a $ which is $5$ for $ n_a ={25,49,55,85,91}$ , indeed number of pairs is $7=12-5$

• similarly if $ 2n = 4 \bmod 6 $ then $ 2n = n_a + n_b $ with $ n_a = -1 \bmod 6 $ and $ n_b = -1 \bmod 6 $ or $ 2n = 3 + n_c $ with $ n_c = 1 \bmod 6 $ it is possible to verify that for each number $ 2n <1084 $ the total number of primes for $ n_a \leq \frac {n} {2} $ considering also $ 3 $ is greater than the total number of composites for $ n_b \geq \frac {n} {2} $ also considering $ n_c $ or that the total number of primes for $ n_b \geq \frac {n} {2} $ is greater than the total number of composites for $ n_a \leq \frac {n} {2} $. For this reason there is always at least one solution.

• finally if $ 2n = 2 \bmod 6 $ then $ 2n = n_a + n_b $ with $ n_a = 1 \bmod 6 $ and $ n_b = 1 \bmod 6 $ or $ 2n = 3 + n_c $ with $ n_c = -1 \bmod 6 $ it is possible to verify that for each number $ 2n <608 $ the total number of primes for $ n_a \leq \frac {n} {2} $ also considering $ 3 $ is greater than the total number of composites for $ n_b \geq \frac {n} {2} $ also considering $ n_c $ or that the total number of primes for $ n_b \geq \frac {n} {2} $ is greater than the total number of composites for $ n_a \leq \frac {n} {2} $. For this reason there is always at least one solution.