A polynomial that factors (poly mod n) but has no roots?

**Pn0yS0ld13r** · July 29th 2008, 12:24 AM

How is it possible for a polynomial to factor (poly mod n) but have no roots? That is no integers x such that $f(x)\equiv0\bmod{n}$ .

Give an example of a polynomial that factors (poly mod n) and prove that it has no roots.

I'm completely stumped.

**Moo** · July 29th 2008, 12:34 AM

Hello,

Originally Posted by Pn0yS0ld13r

How is it possible for a polynomial to factor (poly mod n) but have no roots? That is no integers x such that $f(x)\equiv0\bmod{n}$ .

Give an example of a polynomial that factors (poly mod n) and prove that it has no roots.

I'm completely stumped.

Here are some examples :
If n=2, in $\bmod 2$ , $x^2+x+1$ has no root. (and it's the only one if n=2)

So you can take as an example : $f(x)=(x^2+x+1)^2=x^4+2x^3+3x^2+2x+1 \equiv x^4+x^2+1 \bmod 2$ this factors into $(x^2+x+1) \cdot (x^2+x+1)$ but it has no root $\bmod 2$ .

Check it out : $x \equiv 0~,~1~,~2 \bmod 2$ doesn't give any zero for $x^2+x+1$