# Is $x^{p}+1 \in Z_{p}[x]$ irreducible or reducible? [closed]

If $p$ is a prime number and $x^{p}+1 \in \mathbb Z_{p}[x]$. Show that $x^{p}+1$ isn't irreducible in $\mathbb Z_{p}[x]$. I need know how can show that, is for a homework that I send tomorrow to the University. Only I know that I need use the little theorem of Fermat.

-

## closed as off-topic by TMM, Matthew Conroy, Michael Albanese, YACP, azimutJan 1 at 11:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TMM, Matthew Conroy, Michael Albanese, Community, azimut
If this question can be reworded to fit the rules in the help center, please edit the question.

Hint: it is $\,f(x) - f(-1)\,$ so has root $\,x - ?\,$ by the Factor Theorem. –  Bill Dubuque Jan 1 at 1:41
$(x+1)^p\equiv x^p+1\pmod{p}$ by Binomial-theorem. –  Siddharth Prasad Jan 1 at 1:41
Thanks, but I dont know how I can show that. Plz help –  Vader4891 Jan 1 at 1:59

$x^p+1 = (x+1)^p$ in characteristic $p$ :)

-
Thanks, but I dont know how I can show that. Plz help –  Vader4891 Jan 1 at 1:47
Show $p\mid \binom{p}{k}$ for $1\leq k \leq p-1$. –  Pedro Tamaroff Jan 1 at 2:05
why I need show that? –  Vader4891 Jan 1 at 2:21
You don't really need that actually, it is enough to just show that $-1$ is a root, which will show that the polynomial is reducible (but not give a complete factorisation). –  fkraiem Jan 1 at 2:24
$-1$ is a root because $(x+1)^{p}=0$ if $x=-1$. Thanks to you –  Vader4891 Jan 1 at 2:39

$f=X^p+1$ $$f(-1)=(-1)^p+1=(-1)^{p-1}(-1)+1=1\cdot(-1)+1=-1+1=0$$

$=>(X+1)$ divide $X^p+1$

-

The suggestions thus far are very good. A simply way of solving this without use of the binomial theorem is we know $x^p+1$ is reducible if it has a root in $\mathbb{Z}_p$. For any prime $p$, $-1=1 \in \mathbb{Z}_p$

(Technically this is not so. However, this simplifies the matter and it is simple to find the correct corresponding positive element in $\mathbb{Z}_p$ for $-1$ as $\mathbb{Z}_p \cong \mathbb{Z}/\langle p\rangle$. I simply find it cleaner to use $-1$.).

In any case, if $p$ is an odd prime, then notice when $x=-1$, we have $$x^p+1=(-1)^p+1=-1+1=0$$ so that $-1$ is a root. If $p$ is an even prime (that is $p=2$), we are looking a the ring $\mathbb{Z}_2$, then notice $x=1$ yields $$x^p+1=(-1)^2+1=1+1=0$$ in $\mathbb{Z}_2$ so that again $x^p+1$ has a root and is reducible.

-