Using suitably defined matrices $A$ and $B$, Goldbach's conjecture can be written as:
$$(AB)^2 = 0 \to (AB)=0$$
Considering a matrix $C=AB$, this can be written as:
$$ C^2 = 0 \to C = 0$$
which is an assertion of a zero product property.
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My question is this: What does this imply for the conjecture, given that matrix algebra does not possess the zero product property?
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To see that these are, in fact, statements of Goldbach's conjecture, let n be an even integer and define $n \times n, (0,1)$ matrices $A$ and $B$ with indices ranging from $0$ to $n-1$ as follows:
$$ A=(a_{ij}) \text{ where } a_{ij}=1 \text{ iff } i=j \text{ and }i \text{ is prime }$$ $$ B=(b_{ij}) \text{ where } b_{ij}=1 \text{ iff } i+j\equiv 0\pmod n$$
$A$ is a diagonal matrix with elements equal to $1$ iff its index is prime.
$B$ is a permutation matrix that, when pre and post multiplying $A$ as $BAB$, has the effect of taking the diagonal elements of $A$ from $a_{ij}$ to $a_{n-i,n-j}$ (with the exception of $a_{00}$ because $b_{00}$ is a fixed point).
When $n = 2, A \text{ and } AB$ are both zero while for all $n > 2$, $A \text{ and }AB$ are both nonzero. Equivalently, if $n$ is even and $AB = 0$ then $n$ must be $2$.
The product $A(BAB) = (AB)^2$ will have diagonal elements equal to $1$ iff the indices $i$ and $n - i$ are both prime, and, whenever $n$ is not the sum of two primes, $(AB)^2 = 0$.
With $A$ and $B$ defined as above, the statement:
$$(AB)^2 = 0 \to (AB) = 0$$
or, with $C=AB$:
$$C^2=0 \to C=0$$
is equivalent to:
$$ \text{If $n$ is even and not the sum of two primes, then $n$ is } 2.$$
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To help visualization, the matrices $A$ and $B$ are written here for $n=6$:
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$$A = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$
And the product $(AB)^2$:
$$(AB)^2=A(BAB) = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$
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In the last matrix above, the diagonal element with indices $i=j=3$ is equal to $1$ because $6=3+3$.
In general, if we denote the product by $(AB)^2$ by $G=(g_{ij})$, then
$$g_{ij}=1 \text{ iff } i=j \text{ and } i \text{ and } n-i \text{ are both prime.}$$