Find all zeros of $5x^3 + 4x^2 + 11x + 9 \in \mathbb{Z}_{12}[x]$

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I want to find all of the zeros of the polynomial $5x^3 + 4x^2 + 11x + 9 \in \mathbb{Z}_{12}[x]$. I know that one simple way would be to check every value in the set $x=\{0,1,...,11\}$, and when I did this I found that there were no zeros.

However, is there a more elegant way to show this? I was thinking that I could try to show the polynomial is irreducible, and then use that to help show there are no roots, but couldn't see a good way to show irreducibility in $\mathbb{Z}_{12}[x]$

Thanks for any help you might have!

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2 Answers

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If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.

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There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.

By one of Euler's theorems $a^{\phi(n)}=1\ (\mod n)$, where $\phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.

So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.

$\phi(4)=2$.

$\phi(3)=2$.

We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.

In mod 4, the polynomial becomes :$$5x^3+4x^2+11x+9$$$$(4+1)x^{(\phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$$$x-x+1=0(\ mod \ 4)$$Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.

Let's try with 3. $\phi(3)=2$$$5x^3+4x^2+11x+9$$$$(3+2)x^{\phi(3)+1}+(3+1)x^{\phi(3)}+(3*4-1)x+3*3$$$$2x+1-x=0 (\ mod \ 3)$$$$x=-1 (\ mod \ 3)$$

So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.

However, suppose we did have a solution in mod 4, that our solutions were:

$x=p (\ mod \ 4)$

$x=q (\ mod \ 3 )$

Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$

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