# A motivating argument for studying determinants of $2\times 2$ matrices with prime entries in relation to Goldbach's conjecture.

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
|     |
|     |
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: 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.

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 $$2n$$.

• I see, you used an elementary row op and that preserves the determinant? – CommutativeAlgebraStudent Aug 19 at 9:45
• Also, I think you meant add the columns! – CommutativeAlgebraStudent Aug 19 at 9:46
• 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. – lulu Aug 19 at 9:49
• 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. – CommutativeAlgebraStudent Aug 19 at 9:51
• I already edited it. I don't have time to look at the other questions now, I'll try to return to them later. – lulu Aug 19 at 9:52

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 $$2n$$.