Symmetry, the Goldbach Conjecture and the Twin Prime Conjecture

The Goldbach Conjecture

Veritasium, a YouTube channel with science, released a video on the Goldbach Conjecture. Because the problem statement is so simple, it’s very tempting to play around with the math. Based on a symmetry in addition, there is an equivalent way of stating the conjecture.

The Conjecture

The original conjecture is stated as

Every even integer greater than $2$ can be written as the sum of two primes.

Symmetric Reformulation

My claim is that an equivalent statement is that

For all integers $k$ greater than $1$, there exist at least one integer $d$, $0 \leq d < k$ such that $k-d$ and $k+d$ are prime.

In other words, the two primes that sum up to $n = 2k$ from the Goldbach Conjecture must be at an equal distance $d$ from $k$, the midpoint of $n$. You can find the proof below.

Justification

The symmetry is not due to the Goldbach Conjecture itself, but it originates from addition. For any integers $a,b,k > 0$ such that $a + b = 2k$, $a$ and $b$ must be of the form $k+d$ and $k-d$ for some $d$. The reformulation of the Goldbach Conjecture just follows from applying this fact to the conjecture.

Relation to the Twin Prime Conjecture

The Twin Prime Conjecture is that

There are infinitely many primes $p$ such that $p+2$ is also prime.

This is equivalent to having an integer $k$ with $d = 1$ such that $k-d$ and $k+d$ are prime. Such a twin prime pair would satisfy the Goldbach Conjecture for this $n = 2k$. At the same time, if the Goldbach Conjecture is true, and one could show that we always have some $n$ where $d = 1$, then this would prove the Twin Prime Conjecture.

Some Plots

The Goldbach Conjecture has been shown to be true up to $4\cdot 10^{18}$ ¹. For example for $n = 16$ we know that it can be written as the sum of two primes by $5 + 11$, or by $3 + 13$. The midpoint, $k$, is $8$. For both pairs of numbers, $k$ is in the middle. $k + 3 = 11$ and $k-3 = 5$.

Likewise, we can rewrite $3$ as $8-5$ and $13$ as $8+5$.

The following plot contains some aggregate statistics for the distance $d$. The maximum and minimum clearly follow the bounds $n/2$ and $0$. Fitting the slope for the average size of $d$ for $2 < n \leq 10^6$ we get $f(n) \approx 0.238579x + -43.772569$. The R-Squared metric is $0.999210$, so it’s a good fit.

The linear relationship in the average is due to the distribution of distances $d$ in its search space. For a given $n$, $d$ must be between $0$ and $n/2$. Plotting all the distances, we can see the distances are roughly evenly distributed. Thus, the average of them is approximately in the middle of the search space, $n/4$, which reflects the $0.2385\dots$ in the approximation.

The diagonal grid-like pattern in the distances are reminiscent of the Ulam Spiral² pattern, so they are likely due to the pattern in prime numbers themselves.

Proof

Let $n = 2k$ beany even postive integer. If $a + b = 2k$, for some integers $a,b > 0$, then $a$ and $b$ can be written as $k + d$ and $k-d$ for some integer $d$. We achieve this by reformulating $a + b = 2k$ as $a-k = k-b$, which we define as $d$.

so $(k+d) + (k-d) = a + b = 2k$ always holds.

Assume that $a$ is greater than or equal to $b$. The two summands $a$ and $b$ are greater than zero by assumption. The larger of the two, $a$, can be at most $2k-1$, which is reached with equality when $b = 1$. Thus, $d = a - k \leq 2k-1-k = k-1$, our upper bound for $d$. The lower bound is reached when $a = b = k$, thus $d = 0$.

This proves that any integer $n = 2k$ and $a,b$ such that $a + b = n$, $a$ and $b$ are of the form $(k-d)$ respectively $(k+d)$, for some $d$, $0 \leq d < k$. $\blacksquare$

Application to Goldbach Conjecture

Let $n$ be an even integer greater than $2$. As $n$ is even, we can write $n=2k$ for some $k$. The lower bound on $n$ translates to $k > 1$. As we can map bijectively from all even $n$ to $k$, proving the statement over all $k$ is equivalent.

If we assume the Goldbach Conjecture to be true for some $n = 2k$, then there exist primes $p_1, p_2$ such that $p_1 + p_2 = 2k$. As proven earlier, they must be of the form $k-d$ and $k + d$. If we on the other hand assume $2k$ can be written as the sum of $k-d$ and $k+d$ which are prime, we have shown the other direction, as $2k$ can be written as the sum of two primes. Thus, the reformulation is equivalent to the original conjecture.