Given the below coefficients, if the Diophantine equation $Axy + Bx + Cy + D = \lfloor\frac{n}{3}\rfloor$ has exactly one solution, then $n$ is prime, otherwise $n$ is composite. In a sense, this equation models primes by formalizing the factorization of $n$. The nonnegative solutions for $y$ uniquely encode the exhaustive set of odd factors of $n$. Here are the assumptions: $n=2x+1$ where $x \in \mathbb{N}$, $d=2a+1$ is the divisor of $n$ corresponding to $y$, where $a=y+b$ if $3 \mid n$, otherwise $a=3y+b$ (thus $d$ is decoded from $y$), and where the following coefficients are used:
$$\begin{array}{c|c|c|} & \text{A} & \text{B} & \text{C} & \text{D} & \text{b} \\ \hline \text{1.a} & 6 & 5 & 2 & 1 & 2 \\ \hline \text{1.b} & 6 & 7 & 2 & 2 & 3 \\ \hline \text{2.a} & 6 & 5 & 4 & 3 & 2 \\ \hline \text{2.b} & 6 & 7 & 4 & 4 & 3 \\ \hline \text{3} & 2 & 5 & 0 & 0 & 2 \\ \hline \end{array}$$
For each $n$ the equation must be solved once or twice in order to complete the list of factors. Values of $n$ of the form $6j-1$, for any integer $j>0$, must use the $1.a$ and $2.b$ coefficients, those of the form $6j+1$ must use the $1.b$ and $2.a$ coefficients, and multiples of 3 must use the case 3 coefficients. For case 3, 3 is not in the solution (but choosing case 3 coefficients implies 3 is a factor). Here are the solutions for $n=99$: 3 $\rightarrow$ 11 and 33, for $n=119$: 1.a $\rightarrow$ 119, 17 and 2.b $\rightarrow$ 7, and for $n=157$: 1.b $\rightarrow$ 157. The value $x=0$ for case-1 coefficients is a trivial solution, since that solution corresponds to $d=n$. No values of $y$ correspond to $d=1$. The first four cases simplify to two equations, both of which may be evaluated for any $n=6j\pm1$, where $z=c \pmod 2$, $n=f(c)=2g+1$, $g=c+\lceil\frac{c}{2}\rceil$, and $c=\lfloor\frac{n}{3}\rfloor$:
$6xy + 5x + 4y + 3 - 2z(y+1) = c$
$6xy + 7x + 2y + 2 + 2z(y+1) = c$
My question is, does this qualify as one of the prime-representing Diophantine equations revealed by Matiyasevich? In either case, please explain what aspects of the above equation meet of violate Matiyasevich's criteria.
A representation of a Diophantine set $S \subset \mathbb{N}$ is a polynomial with integer coefficients
$P(X_1,X_2,\ldots,X_n,Y)$
with the following property: $y \in S$ if and only if there exist $x_1,x_2,\ldots,x_n \in \mathbb{N}$ such that $P(x_1,x_2,\ldots,x_n,y)=0$.
That is all. You need one polynomial in some number of variables, and the coefficients of that polynomial are integers.
For example, the polynomial
$(X_1+2)(X_2+2)-Y$
has a solution with $Y=y$ if, and only if, $y$ is a composite natural number. Therefore the set of composite numbers is a Diophantine set.
If you want to show that the set of primes is Diophantine, you need to exhibit such a polynomial. The equation $Axy+Bx+Cy+D=\lfloor \frac{n}{3} \rfloor$ doesn't qualify since the floor function is not part of the language of polynomials, and since you didn't say what $A,B,C,D$ are: you gave several options which seem to depend on the other variables. A polynomial has specific, fixed coefficients.
There are, indeed, Diophantine equations defining the set of primes, but all known such equations have high degree or a large number of variables (or both). It is extremely unlikely that there is such an equation with just two variables, though I'm not aware that this is a proven result. It is certain that there is no such equation with two variables and degree $2$, as your example attempts (I think) to show.