I want to apply the Chinese remainder theorem to the polynomial equation $ x^{3}+2x+1 \equiv 0 \bmod 15 $ which I have split into two equations $x^{3}+2x+1 \equiv 0 \bmod 3 $ and $x^{3}+2x+1 \equiv 0 \bmod 5$.
Looking at the Chinese remainder theorem the congruence equations are linear but I have seen examples which have applied the theorem to quadratic equations so is it possible to apply it this equation?
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Yes, you may use the chinese remainder theorem.
$x^3+ 2x + 1 \equiv 0 \mod 3$
By Fermat $x^3 \equiv x \mod 3$ and we get the impossible $x^3 + 2x + 1 \equiv x + 2x + 1 \equiv 3x + 1 \equiv 1 \equiv 0\mod 3$. (Or we could have tried them each individually-- there's only three of them.)
so the chinese remainder theorem shows there is no solution.
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As an example where it would have worked. Let's let $x\equiv 6 \mod 15$ and so $x^3 + 2x + 1\equiv 4\mod 15$ so suppose we we asked to solve $x^3 + 2x -3 \equiv 0 \mod 15$.
Then $x^3 + 2x -3 \equiv 0 \mod 3$ yields
$x^3 - x \equiv 0 \mod 3$
$x(x+1)(x-1) \equiv 0 \mod 3$ and any value mod 3 will solve this.
$x^3 + 2x -3 \equiv 0 \mod 5$ yields
$x^3 + 2x +2 \equiv 0 \mod 5$. $x = 0$ is not a solution. So if there is any solution it is relatively prime to $5$.
$x^4 + 2x^2 + 2x \equiv 2x^2 + 2x + 1 \equiv 0 \mod 5$.
$x \equiv \frac {-2 \pm \sqrt {4 - 4*2}}{4} = 2\pm \sqrt {-1+2}\equiv 2 \pm 1$.
So $x \equiv 1,3 \mod 5$.
So solutions are all $1,3,6,8,11,13\mod 15$. All in all, there was probably an easier way... but that was valid
It is perfectly fine to apply the theorem in this situation, any equation mod $15$ is equivalent to the same equation holding simultaneously mod $3$ and mod $5$.
Hint to continue: use Little Fermat's theorem.