# Number of solutions for $n^5 + 2 n^4 + n^3 - 3n + 2$ mod $23^2 = 0$, where $0 \leq n < 23^2$ and $n\in \mathbb{N}$

$0 \leq n < 23^2$ and $n\in \mathbb{N}$
For how many $n$
$n^5 + 2 n^4 + n^3 - 3n + 2$ mod $23^2 = 0$

• You must try to show your attempt. May 7 '16 at 12:50
• I don't even know where to start. It is beyond capability of my math skills. May 7 '16 at 12:53
• @Coincidence : May be you can try solving that for some small number, preferably prime to get some idea.. $23$ is a big number.. try $5$
– user311526
May 7 '16 at 12:58

$$n^5 + 2n^4 + n^3 - 3n + 2 = (n+2)(n^4 + n^2 - 2n + 1) = (n+2)(n^4 - (n-1)^2) = (n+2)(n^2 - n + 1)(n^2 + n - 1)$$
By the quadratic formula, the last two factors cannot split in fields where $\sqrt{-3}$ and $\sqrt{5}$ are not members of the field, respectively. However, the quadratic residues in $\mathbb{Z}/23\mathbb{Z}$ are $1, 4, 9, 16, 2, 13, 3, 18, 12, 8, 6$ (obtained by squaring the first 11 elements.) Neither $-3 = 20$ nor $5$ is in this list, therefore the final two factors are irreducible in $\mathbb{Z}/23\mathbb{Z}$, and do not have roots lying in the field. (This means that they are never divisible by $23$, and by extension are not zero divisors in $\mathbb{Z}/23^2 \mathbb{Z}$.)
As such, the only possibility is for the first factor to be divisible by $23^2$, therefore the polynomial has only one root in $\mathbb{Z}/23^2\mathbb{Z}$.