Proving systems of nonlinear modular equations have no solution I have reason to suspect this system of six nonlinear modular equations has no solution for $2 < x < y < z$ even integers.
$$
\left\{
\begin{aligned}
z(3y+2) \equiv y(3z+2) \equiv 0& \mod x\\
 z(3x+2) \equiv x(3z+2) \equiv 0& \mod y\\
x(3y+2) \equiv y(3x+2) \equiv 0& \mod z
\end{aligned}
\right.
$$
Remove any one and numerous solutions are easy to find, so I can't make the system any smaller.  Having no solution is also consistent with other empirical results.  But naturally I'd like a proof or a counterexample.  Since the moduli are not pairwise coprime, I don't see how the CRT can help.
Using the definition of modulus, I could transform these into a system of nonlinear equations with nine variables:
$$\begin{bmatrix}
    -K_1   &2   &3y\\
    -K_2  & 3z &  2\\
    2   & -K_3   &3x\\
    3z   &-K_4  & 2\\
    2   & 3x   &-K_5\\
    3y   & 2   &-K_6
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0
\end{bmatrix}
$$
where the Ki's are positive constants.  But this doesn't seem to get me anywhere and may actually make the problem harder.
Any hope of proving something like this? This is research, but I'm a computer scientist by training, not a mathematician.  Abstract algebra, discrete math, number theory, linear algebra etc either self-taught or learned back in the Dark Ages by candlelight.
--BF
 A: Proof outline


*

*We first simplify the system into (almost) coprime modulus using only $3$ of the $6$ equations (ignoring the $\equiv 0$ part).  

*This will allow us to derive 2 general classes of solutions.  

*We will then use a fourth equation $x(3z+2)\equiv 0 \pmod y$ to show both are not feasible, concluding the proof that there are no solutions.  



Since $2<x<y<z$ are even integers, we let $x=2r,y=2s,z=2t$ and part of the system of equations (using the $\equiv 0$ part later) becomes
$$
\begin{align*}
2z-2y&\equiv 0 \pmod x &\implies 2t-2s &\equiv 0 \pmod r\\
2z - 2x &\equiv 0 \pmod y &\implies 2t-2r &\equiv 0 \pmod s\\
2y - 2x &\equiv 0 \pmod z &\implies 2s-2r &\equiv 0 \pmod t
\end{align*}
$$
Let $d = \gcd(s,t)$ and write $s=bd,t=cd$ so that $\gcd(b,c)=1$. Then from
$$
2s - 2r= 0 \pmod t,
$$
we get
$$
2bd - 2r \equiv 0 \pmod{cd}
$$
So that $d$ divides $2r$. Hence we let $2r = ad$. We now simplify the system of equations:
$$
\begin{align*}
2t-2s &\equiv 0 \pmod r &\implies 4t-4s &\equiv 0 \pmod{2r}&\implies 4cd-4bd &\equiv 0 \pmod{ad} &\implies 4c-4b&\equiv 0\pmod a\\
2t-2r &\equiv 0 \pmod s &\implies 2cd-ad &\equiv 0 \pmod{bd} &\implies 2c-a &\equiv 0 \pmod b\\
2s-2r &\equiv 0 \pmod t &\implies 2bd-ad &\equiv 0\pmod{cd} &\implies 2b-a &\equiv 0 \pmod c
\end{align*}
$$
Hence we get a new system
$$
\begin{align*}
4c-4b &\equiv 0 \pmod{a}\\
2c-a &\equiv 0 \pmod{b}\\
2b-a &\equiv 0 \pmod{c}
\end{align*}
$$

Next we prove a lemma that bounds the values of $a,b,c$:

Lemma. Any solution to the system must satisfy
  $$
1\leq a,b,c \leq 9
$$

Proof. From the system of equations:
$$
\begin{align*}
2c - a &\equiv 0 \pmod b &\implies 2b + 2c - a &\equiv 0 \pmod b\\
2b - a &\equiv 0 \pmod c &\implies 2b + 2c - a &\equiv 0 \pmod c
\end{align*}
$$
Since $\gcd(b,c)=1$, by CRT we have
$$
2b+2c - a \equiv 0 \pmod{bc}
$$
Since $2b = 2s/d > 2r/d = a$, this means $2b+2c-a > 0$. Therefore we obtain a bound of $bc$:
$$
bc \leq 2b+2c - a < 2b+2c
$$
If $3 \leq b < c$, then
$$
(b-2)c < 2b \implies c < (2b)/(b-2) = 2 + 4/(b-2) \leq 2+4 = 6
$$
giving us a bound of $c \leq 5$. Similarly,
$$
(c-2)b < 2c \implies b < (2c)/(c-2) = 2 + 4/(c-2) \leq 2+4 = 6
$$
Hence we get $b,c \leq 5$. Using $2b > a$ then bounds $a,b,c$ as $a,b,c \leq 9$.  
For the remaining case, $b=1$ or $b=2$. If $b=1$ then $2b>a$ forces $a=1$, so the system reduces to
$$
2-1 \equiv 0 \pmod{c}
$$
This forces $c=1$, contradicting $c>b$. Alternatively, if $b=2$ the system becomes
$$
\begin{align*}
4c-8 &\equiv 0 \pmod{a}\\
2c-a &\equiv 0 \pmod{2}\\
4-a &\equiv 0 \pmod{c}
\end{align*}
$$
So we see that $2$ divides $a$. Since $1\leq a < 2b = 4$, this can only be $a=2$. But now
$$
4-2 \equiv 0 \pmod c
$$
forces $c=1,2$, again contradicting $c>b$. This means $b\leq 2$ is not possible, therefore the previous bound
$$
1\leq a,b,c \leq 9
$$
is the only possible one and we are done.
$$
\tag*{$\square$}
$$

Now a brute force search of $1\leq a,b,c \leq 9$, conditioned on $a/2<b<c$ and $\gcd(b,c)=1$ shows that the only solutions are
$$
(a,b,c) = (1,3,5), (2,3,4)
$$
which corresponds to
$$
(x,y) = (2r,2s) = (ad,2bd) = (d,6d), (2d,6d)
$$
($z$ doesn't matter) They must satisfy one of the original equations
$$
x(3z+2) \equiv 0 \pmod y
$$
Hence we must have
$$
\begin{align*}
(d)(3z+2) \equiv 0 \pmod{6d} \implies 3z+2 \equiv 0 \pmod 6\\
(2d)(3z+2) \equiv 0 \pmod{6d} \implies 2(3z+2) \equiv 0\pmod 6
\end{align*}
$$
This is impossible $\pmod 3$, therefore there are no solutions and we are done.
A: Edit: Made a mistake, this is wrong 
Let us isolate two of the equations:
$$
\begin{align*}
z(3y+2) &\equiv 0 \pmod x\\
z(3x+2) &\equiv 0 \pmod y
\end{align*}
$$
From
$$
z(3y+2)\equiv 0 \pmod x,
$$
This means $x$ divides $z(3y+2)$. Since these are positive numbers,
$$
x \leq 3yz+2z
$$
Similarly, from
$$
z(3x+2)\equiv 0 \pmod y
$$
we get
$$
y \leq 3x z+2z
$$
Now subtracting latter from former:
$$
y-x \leq 3z(x-y)
$$
Since $x < y$, LHS is greater than zero. However since $x<y$ and $0<z$, RHS is less than zero. Therefore this is impossible.
