# Does there exist a polynomial it does not have any integer root but has at least one root in $\Bbb Z_n$ , $\forall n \in \Bbb Z$?

Does there exist a polynomial $f \in \Bbb Z[X]$ such that $f$ does not have any integer root but $f$ has at least one root in $\Bbb Z_n$ , $\forall n \in \Bbb Z$ ( while considering $f$ as a polynomial over $\Bbb Z_n$ ) ?

If there exists such a polynomial then we need to Construct it, otherwise a general proof will suffice.

I am trying to work it out with the amount of Ring theory I know but so far haven't been able to do anything!

• Since at least one of $2,3,6$ must always be a square $\pmod p$ for any prime $p$ the polynomial $(x^2-2)(x^2-3)(x^2-6)$ will always have a root $\pmod p$. A little more effort shows that in fact it has a root $\pmod n$ for all $n\in \mathbb N$. – lulu Aug 13 '18 at 14:37
• Possible duplicate of math.stackexchange.com/questions/2863816/… – Daniel Schepler Aug 14 '18 at 0:55

$f(x)=(x^2-13)(x^2-17)(x^2-221)$ works. First note that $17*13=221$. Therefore, for prime $n$, one has $$\left ( \dfrac{221}{n} \right )=\left ( \dfrac{13}{n} \right ) * \left ( \dfrac{17}{n} \right ).$$ In particular, not all three numbers 13, 17, and 221 can be quadratic non-residues modulo $n$ at the same time. So $f(x)$ has a root modulo every prime $n$.
For composite $n$, suppose on the contrary, that $f(x)=0$ has no roots modulo $n$ and we derive a contradiction. In other words, suppose 13, 17, and 221 are quadratic non-residues modulo $n$. It follows that a prime power factor of $n$ exists, such as $p^k$ such that 13, 17, and 221 are quadratic non-residues modulo $p^k$. If $p$ is odd, a number is a quadratic residue modulo $p$ if and only if it is a quadratic residue modulo every power of $p$, so we are done in this case. If $p=2$, then 17 is a quadratic residue modulo $2^k$, a contradiction.
• It might also be worth giving special attention to the cases $p=13$, $p=17$ - and fleshing out the argument for $p=2$ (e.g. if $n$ is a root of $n^2 + n - 4$ in $\mathbb{Z}_{2^k}$ then $2n+1$ is a square root of 17 in $\mathbb{Z}_{2^k}$ and Hensel's lemma applies to the polynomial $n^2 + n - 4$ - but not to the polynomial $m^2 - 17$ which is why we use the other one.) – Daniel Schepler Aug 14 '18 at 1:02