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Solve $x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod {(4\cdot3)}$


Work:

Let $P(x) = x^{10} +4x^3+3x+4 \equiv 0 \pmod 2$ and $12=4\cdot3=2^2\cdot3$

We have $P(x) \equiv0 \pmod {2^2}$ and $P(x)\equiv0\pmod3$

For $\pmod {2^2}$,

$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 2$

Try $[0],[1]$,

$[0]$ works since $P(0) =4 \equiv0\pmod2$, say $a_1 =0$

$[0]$ lifts to a unique solution to a class that $[a_2] \in \mathbb{Z}_{2^2}$ so that $P(a_2) \equiv 0 \pmod {2^2}$

Solving $P'(a_1)\cdot x =3\cdot{x}\equiv1\pmod2$, $\overline{P'(a_1)}=x=1$

any representative in class of

$a_2 = a_1 -\overline{P'(a_1)}\cdot {P'(a_1)}=3-1\cdot3=0\equiv{0}\pmod{2^2}$

$0\pmod{2^2}$ the only solution to $P(x)\equiv{0}\pmod{2^2}$

For $\pmod3$,

$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 3$

Try $[0],[1],[2]$, $[1]$ works. Say $a_2=1$

$P(1)$ is a solution to $\pmod3$

and how do I combine 2 cases??

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If you want $n \equiv 1 \pmod 3$ and $n \equiv 0 \pmod 4$ you can just try the multiples of $4$ until you find one that works. In this case it is $4$. So $P(4)\equiv 0 \pmod {12}$. Here is a check from Alpha

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  • $\begingroup$ mutliply 4 with?? $\endgroup$ – Paul Mar 12 '13 at 4:03
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    $\begingroup$ @Paul: modulo $12$ there are only three multiples of $4: 0,4,8$ One of them will have each remainder mod $3$, so you want to find the one that is $1 \pmod 3$. This is a case of the Chinese Remainder Theorem. $\endgroup$ – Ross Millikan Mar 12 '13 at 4:07

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