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I am trying to solve the following problem:

How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c \in \mathbb{Z_{18}}$, have?
($\mathbb{Z_{18}} = \{0, 1, 2, \ldots, 17\}$)

Obviously, they can not have more than 18 roots.

If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.

I tried this...

Let $x_1$ be a root of $p(x)$. Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$

I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $\mathbb{Z_{18}}$.

Is any of this in the right direction? I could use some help.

And is there a more general way I could describe possible roots of polynomials from that ring?

Blue
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Coupeau
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1 Answers1

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[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]

The way I would approach this is using the Chinese remainder theorem. Since $\mathbb{Z}_{18}\cong \mathbb{Z}_2\times\mathbb{Z}_9$, we can think about the problem separately in $\mathbb{Z}_2$ and $\mathbb{Z}_9$.

Now $\mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.

Notice now that any quadratic $ax^2+bx+c$ over $\mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $a\neq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $\mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $\mathbb{Z}_{18}$ with all $18$ elements as roots!

Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $\mathbb{Z}_{18}$.

Eric Wofsey
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