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

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3
  • $\begingroup$ You must try to show your attempt. $\endgroup$
    – N.S.JOHN
    May 7 '16 at 12:50
  • $\begingroup$ I don't even know where to start. It is beyond capability of my math skills. $\endgroup$ May 7 '16 at 12:53
  • $\begingroup$ @Coincidence : May be you can try solving that for some small number, preferably prime to get some idea.. $23$ is a big number.. try $5$ $\endgroup$
    – user311526
    May 7 '16 at 12:58
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$$ n^5 + 2n^4 + n^3 - 3n + 2 = (n+2)(n^4 + n^2 - 2n + 1) = (n+2)(n^4 - (n-1)^2) = (n+2)(n^2 - n + 1)(n^2 + n - 1) $$

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} $.

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