Give an example of a polynomial $x^2 + ax + b \in R[x]$, where $R = \mathbb{Z}$ / $4\mathbb{Z}$, which has 3 distinct roots in $R$.

My immediate thought is that there is no such polynomial because the degree of of the polynomial is less than the number of roots we're looking for. Is that right or naive?

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    $\begingroup$ False,Note that $ \mathbb Z_4$ is not a field. $\endgroup$ Oct 18, 2015 at 13:47
  • $\begingroup$ Thanks. I have wrote down all combinations of the polynomial with $a,b = 0,1,2,3$ and can only find equations with 2 roots. What am I missing? $\endgroup$ Oct 18, 2015 at 13:50
  • $\begingroup$ Are you sure the polynomial is supposed to be monic? $\endgroup$
    – Arnaud D.
    Oct 18, 2015 at 14:03
  • $\begingroup$ Yes it is monic. So if there is no such polynomial, is the only reason I can give is that I've exhausted the possible polynomials and not found one? $\endgroup$ Oct 18, 2015 at 14:26
  • $\begingroup$ You could talk about the values that $x^2$ and $ax$ take, and what that means about what you get when you add the two together and pick two values to be the same. But an exhaustive search is probably at least as good a reason. $\endgroup$
    – user24142
    Oct 18, 2015 at 16:08

1 Answer 1


In general we can only conclude that the number of roots of a nonzero polynomial $p$ over a ring $R$ is $\leq \deg p$ when $R$ is a field.

Example We have $x^3 = x \bmod 6$ for all $x$ modulo $6$, so the (cubic) polynomial $x^3 - x$ has $6$ roots in $\Bbb Z / 6 \Bbb Z$.

Now, $\Bbb Z / 4 \Bbb Z$ contains a zero divisor, namely, $[2]$, and so is not a field. That said, an exhaustive search shows that no polynomial $x^2 + a x + b$ modulo $4$ has $\geq 3$ distinct roots, but there are two such polynomials, namely $x^2$ and $x^2 + 2 x + 1 = (x + 1)^2$ with two roots of multiplicity two each, and hence total multiplicity larger than the degree of the polynomial. There is a unique quadratic polynomial with $> 2$ distinct roots modulo $4$, namely, $2 x^2 + 2 x = 2 (x + 1) x$.

  • $\begingroup$ Thank you for the examples. So since there is no such polynomial for my question, is the only reason I can give is that I've checked all possible polynomials and they have $<3$ distinct roots? $\endgroup$ Oct 18, 2015 at 14:29
  • $\begingroup$ You're welcome. I think one could construct a general argument using that $x^2 = 0, 1 \bmod 4$. We can then handle these cases separately, in both cases this amounts to finding the number of roots a linear polynomial can have modulo $4$. $\endgroup$ Oct 18, 2015 at 14:41
  • $\begingroup$ I have another small question if you would be able to help, to save writing another post if the answer is trivial? Can you show that $a^2 + b^2 = 1$ has infinitely many solutions? I don't know if it is true. $\endgroup$ Oct 18, 2015 at 15:25
  • $\begingroup$ The answer is easy but not wholly trivial (I assume you're looking for solutions over $\Bbb Q$, in which case the claim is true). Anyway, such an unrelated question certainly belongs in its own post, but probably it's already been asked here. $\endgroup$ Oct 18, 2015 at 16:34

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