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Solve (if possible)the congruence involving polynomial


My work:

Since $45=3^2\cdot5$, we have

$x^3+2x-3\equiv{0}\mod(3)$ and $x^3+2x-3\equiv{0}\pmod 5$

In $\mathbb{Z}_3$,

$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 3$

We have $[0],[1],[2]$

In $\mathbb{Z}_5$,

$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 5$

We have $[0],[1],[2],[3],[4]$

So does it mean I have 3 solutions, $[0],[1],[2]$???

And how should I finish it up?

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To use your method, work modulo $9$ and $5$, and then put things together using the Chinese Remainder Theorem. Note for example that modulo $5$ there are two solutions. – André Nicolas Feb 13 '13 at 20:34
@Paul You can manually check that $0$ (or rather $[0\textbf{]}$) isn't a root. – Git Gud Feb 13 '13 at 20:36
up vote 2 down vote accepted

Here is one way to find the roots $\rm\,mod\ 9.\:$ Since $\rm\,9\mid(x-1)(x^2+x+3),\:$ by unique factorization, either $\rm\,9\mid x-1,\,\ 9\mid x^2+x+3,\:$ or $\rm\:3\mid x-1,\,x^2+x+3.\:$ The last case is impossible since $\rm\:3\mid x-1\,\Rightarrow\,mod\ 3\!:\ x\equiv 1\,\Rightarrow\, x^2+x+3\equiv 2\not\equiv 0.\:$ In the second case we have $\rm\:mod\ 9\!:\ x^2+x+3\equiv x^2+x-6 = (x-2)(x+3).\:$ Again, either $\rm\,9\mid x-2,\,\ 9\mid x+3,\,$ or $\rm\,3\mid x-2,\,x+3.\:$ The latter is impossible, else $\rm\,2\equiv -3\,\ (mod\ 3).\:$ So this case yields two roots $\rm\:x\equiv 2,\, -3.\:$ Finally, the first case $\rm\,9\mid x-1\,$ yields the root $\rm\:x\equiv 1\,\ (mod\ 9).$

$\rm mod\ 5\!:\ x^2+x+3\equiv x^2+x-2 \equiv (x-1)(x+2),\:$ so there are two roots $\rm\:x\equiv 1,-2\,\ (mod\ 5).$

Now employ the CRT (Chinese Remainder Theorem) to combine the solutions modulo $9$ and $5$ to obtain all of the solutions modulo $\,\rm lcm(9,5) = 45.$

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