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Let $\Bbb{P}$ be the set of odd primes. Let $X_n$ for $n \geq 3$ be the Goldbach solution set $X_n = \{(p,q) \in \Bbb{P}\times \Bbb{P} : 2n = p + q \}$.

Suppose that for combinatorial reasons we are concerned with the size of the solution set, since for example it's not always $1$: $13 + 3 = 11 + 5 = 16$.

Question 1. Does $n = 6$ appear to be the largest integer such that $|X_n| = 1$, namely $5 + 7 = 12$ (and that's all)?

Now consider the case of $2n = 16$. We have that $\det \begin{pmatrix} 13 & 3 \\ 11 & 5 \end{pmatrix} = 2 \cdot 16$ and this seems to happen all time, i.e. whenever there are at least two solutions $(p,q), (r,s) \in X_n$ then their determinant is a multiple of $2n$.

Question 2. Can you think of a reason why this might be the case?

Define the standard form of the matrix for two distinct solutions $(a,b), (c,d)$ where $a \lt c \lt d \lt b$ to be $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$

Intuitively speaking, this gives the least, positive multiple of $2n$ as a result.


Next, what is the relationship between all the solutions? Can we have an equation in $2 \times 2$ determinants?

I think this also is true. Take for example the case of $|X_{23}| = 4$ i.e. $X_{23} = \{ (23,23), (41,5), (43,3), (29, 17)\}$

Then there are a total of 6 relationships (matrices) by looking at a square graph and filling in the two diagonal relationships as well.

They are, in standard form:

$$ \det \begin{pmatrix} 43 & 3 \\ 23 & 23 \end{pmatrix} = 20 \cdot 46 \\ \det \begin{pmatrix} 43 & 3 \\ 41 & 5 \end{pmatrix} = 2 \cdot 46 \\ \det \begin{pmatrix} 43 & 3 \\ 29 & 17 \end{pmatrix} = 14 \cdot 46 \\ \det \begin{pmatrix} 41 & 5 \\ 23 & 23 \end{pmatrix} = 18 \cdot 46 \\ \det \begin{pmatrix} 41 & 5 \\ 29 & 17 \end{pmatrix} = 12 \cdot 46 \\ \det \begin{pmatrix} 29 & 17 \\ 23 & 23 \end{pmatrix} = 6 \cdot 46 $$

Now see that if we split their values up in a certain way (we can just look at the mulitple scalar), we have:

$18 + 12 + 6 = 2 + 14 + 20$

And of course you can move all to one side to get an alternating-sign summation that equals zero. But how should we assign the coefficient $\pm 1$? I did some brain storming. And we should do it like so:

Assign a letter $A, B, C, D$ to each solution, simply do it in the order given i.e: $X_{23} = \{A:(23,23), B:(41,5), C:(43,3), D:(29, 17)\}$.

Then draw a square with labels $A,B,C,D$ at the vertices:

A---- B
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C-----D

Now, the matrices have an orientation, e.g. $\begin{pmatrix} A \\ B \end{pmatrix}$ will correspond to an edge direciton of $A \to B$, so fill in the square along with diagonals and edge directions according to the $6$ oriented matrices presented above in the vertical list of determinants. You get a picture that looks like this:

Goldbach's conjecture simplex orientation

Notice that the $+1$ sign lies on a "commuting triangle"-looking face, and the $-1$ sign applies only to the odd man out vertex $C$ which points to all other vertices. $C$ is the only vertex that points to all other vertices, in other words and its outgoing edges (relationships) get the opposite sign.

Putting this all together we get:

$$ (B \to A) + (B \to D) + (D \to A) - (C \to A) - (C \to B) - (C\to D) = 0\\ 18 + 12 + 6 - 20 - 2 - 14 = 0 $$

Question 3. What area of mathematics does the orientation assignment most closely resemble, is it simplicial complexes or can we get even more specific? I would like to read up on whatever you specify.

Question 4. Can you immediately see any counter-example that would rule out the above ideas?

Question 5. If not 4., is there immediately an obvious proof of the above things? Namely that the determinant of a relationship is always a multiple of $2n$ and that the alternating sum of all possible relationship determinants in standard form sums to $0$?


Not only do we have the alternating sum of the tetrahedron, but each of its faces can also be assigned an $\pm 1$ sign and independently sum to zero, for example $14 + 6 - 20 = 0$. I'm still unsure how this (completely separate and different) assignment relates to the larger assignment in the tetrahedron.

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For Question $2)$:
Writing your matrix as $$ \begin{bmatrix} r & s \\ p & q \end{bmatrix} $$

With $r+s=2n=p+q$

Add the first column to the second. Noting that $p+q=2n=r+s$ we see that your determinant is the same as the determinant of $$ \begin{bmatrix} r & 2n \\ p & 2n \end{bmatrix} $$

which is obviously divisible by $2n$.

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  • $\begingroup$ I see, you used an elementary row op and that preserves the determinant? $\endgroup$ – CommutativeAlgebraStudent Aug 19 at 9:45
  • $\begingroup$ Also, I think you meant add the columns! $\endgroup$ – CommutativeAlgebraStudent Aug 19 at 9:46
  • $\begingroup$ Invariance under elementary row operations is one of the key properties of determinants. For $2\times 2$ matrices, of course, one could just write it out. And, yes...for whatever reason I took the transpose of your matrix (which also doesn't change the determinant). I can edit to fix that, if it matters. $\endgroup$ – lulu Aug 19 at 9:49
  • $\begingroup$ no, your post is fine. Thank you! What about the other questions such as assigning orientation. How should we accomplish that in general? I am a simplicially new person. $\endgroup$ – CommutativeAlgebraStudent Aug 19 at 9:51
  • $\begingroup$ I already edited it. I don't have time to look at the other questions now, I'll try to return to them later. $\endgroup$ – lulu Aug 19 at 9:52
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Question 2. The vector $(1,1)$ is an eigenvector with eigenvalue $2n$. The sum of eigenvalues is the trace. So the second eigenvalue $b$ is an integer. Hence the matrix is aimilar to $$\left(\begin{array}{cc} 2n & a \\ 0 & b \end{array}\right)$$ Hence the determinant of the matrix is $2nb$, a multiple of $2n$.

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