Chinese remainder Theorem in polynomials in one variable. In a question I was  answering, I needed to solve these congruences to proceed, and find the least $k<1000$
$$2k+k^2 = 0 \pmod 3$$
$$ 2k^3 + 6k = 0 \pmod 7$$
$$ k = 0 \pmod 2$$
My try:
due to the third statement
$$ k = 2j$$ for  some integer $j$,
due to the second statement
$$ k  = 7x ; k= 7x+2$$ for some integer $k$
due to the first statement
$$ k = 3y+1; k = 3y$$
can I just exhaust all of the $4$ possible cases due to these statements and use the Chinese remainder theorem? I don't know if I'm doing it right or not.
 A: As pointed out in the comments, the solutions for the second congruence are $k\equiv0,2,5\bmod7$. Now combine the congruences one-by-one: the first and third give $k\equiv0,4\bmod6$, which when combined with the second gives
$$k\equiv0,12,16,28,30,40\bmod42$$
So the least positive $k$ is $12$, and the greatest $k<1000$ is $996$.
A: $\bmod 2\!:\,\ k\equiv 0\ \ $  (so the summand for this in the CRT formula will be $0$ and can be ignored)
$\bmod 3\!:\,\ 0 \equiv k(k+2)\iff \color{#0a0}{k\equiv 0,1},\,$ call them $\,\color{#0a0}{k \equiv a}$
$\bmod 7\!:\,\ 0\equiv k(2k^2-1)\iff \,k\equiv\color{#c00} 0\,$ or $\,k^2\equiv 1/2\equiv 8/2\iff k\equiv \color{#c00}{\pm 2},\,$ call them $\,k\equiv\color{#c00} b$
By $ $ CRT: $\,\ \begin{align}&k\equiv 0\!\!\!\pmod{\!2}\\&k\equiv \color{#0a0}a\!\!\!\pmod{\!3}\\&k\equiv \color{#c00}b\!\!\!\pmod{7}\end{align}\iff  k \equiv -14\,\color{#0a0} a - 6\,\color{#c00} b\pmod{\!42}\ $ 
Next we need to substitute all possible values of $a,b$ into the above CRT formula.
So  $\ \ \ \color{#0a0}{a\equiv 0}\,\Rightarrow\, k \equiv -6\,\color{#c00}b \equiv -6[\color{#c00}{0,-2,2}]\equiv 0,12,30\,\pmod{\!42}$
and $\,\color{#0a0}{a\equiv 1}\,\Rightarrow\, k\,\equiv\, -14-6b\, \equiv\, -14\,+\, [0,12,30]\equiv 28,40,16$
A: Least non-negative $k$ is $0$. If you are looking for non-zero solutions then, essentially $k = 2j$; $j = 7i+1, 7i-1$ or $7i$; $i = 3b, 3b+1$;
