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What are some equivalent statements of (strong) Goldbach Conjecture ?

We all know that Riemann Hypothesis has some interesting equivalent statements. My favorites are involved with Mertens function, error terms of Prime Number Theorem, and Farey sequences. Those equivalent statements do not use Riemann Zeta function directly, but provide additional insights about Riemann Hypothesis from very different angles.

What are some equivalent statements of (strong) Goldbach Conjecture ? to shed lights on this problems from different angles ?

david
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9 Answers9

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For every integer $n \geq 2$ there exist integers $k, p$ and $q$ with $0 \leq k \leq n-2$ and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)

Loreno Heer
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For every integer $n \geq 1$ there exists primes $p$ and $q$ such that $\varphi(p)+\varphi(q)=2n$

where $\varphi$ is Euler's Totient function .

Pedja
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I conjectured this:

Every integer $n>3$ is halfway between $2$ primes.

See the proof here .

By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!

François Huppé
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    That the RH is equivalent to $\pi(x)-Li(x)= O(x^{1/2+\epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture. – reuns Jan 05 '19 at 02:44
  • For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine... – François Huppé Jan 05 '19 at 03:30
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    $p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = \frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$). – reuns Jan 05 '19 at 04:03
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"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"

... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"

Perspectiva8
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Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), n\ge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.

It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.

It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.

Keith Backman
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Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3\leq p_i<m<p_j<2m$, let $n=\prod{p_j}$. Let $c_i=2m-p_i$, and $a=\prod{c_i}$. Then $\left(\frac{a}{n}\right)=0$ where $\left(\frac{a}{n}\right)$ is the Jacobi symbol.

Keith Backman
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Just some of the equivalents I've come across in my head etc.:

  • every natural number n, greater than 3 is equidistant 2 odd primes ( arguably distinct)
  • Every natural number n, greater than 3, has between n and 2n-2 a prime, that's congruent to the additive inverse of an odd prime below n.
  • every product of odd primes, is a difference of squares.
  • forall primes r greater than 3, either r+r is the only goldbach partition of 2r, or r is in an arithmetic progression of primes,of length 3 or more.
  • The sum of the squares of any 2 odd primes with at least 1 distinct from 3, is always twice a sum of squares, with 1 base greater than 3.
  • Any two odd primes with phi values that sum to more than 6, have a pair of primes equidistant those phi function values.
  • etc.
  • How did you derive that "every product of odd primes, is a difference of squares" is equivalent to the (strong) Goldbach conjecture? Any proof? Because do remark that for every odd integer $2n+1$, we have $2n+1=(n+1)^2−n^2$. Thus not only can every odd prime be represented as the difference of two squares, but every odd integer can be written as the difference of two squares. – Ok-Virus2237 Mar 06 '24 at 20:12
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Goldbug's Conjecture: The set of Goldbug Numbers is finite.

Goldbug
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$$\sum_{_{3\leq p\leq2n-3}}\omega\left(2n-p\right)>\sum_{_{3\leq p\leq2n-3}}\pi_{p,b}\left(2n-3p\right)$$

where $p$ is prime, $\omega$ is the prime omega function, $\pi_{a,b}$ is the modular prime counting function and $b=\left(2n-3p\right)\mod p$.

François Huppé
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