Chinese remainder Theorem in polynomials in one variable.

Question

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.

Answer 1

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 $996$.

Answer 2

$\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 $\,\color{#0a0}{a\equiv 1}\,\Rightarrow\, k\,\equiv\, -14-6b\, \equiv\, -14\,+\, [0,12,30]\equiv 28,40,16$

Answer 3

Least non-negative $k$ is $0$. If you are looking for non-zero solutions then, essentially $k = 2j$; $j = 7i+1, 7i-1$ or $7i$; $i = 3b, 3b+1$;