# Thread: Polynomials which factors but have no roots?

1. ## Polynomials which factors but have no roots?

Hello all,

Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

Any help is appreciated!

2. Originally Posted by 1337h4x
Hello all,

Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

Any help is appreciated!
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$

Now consider this polynomial modulo $2$.

3. Originally Posted by chiph588@
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$

Now consider this polynomial modulo $2$.
Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

4. Originally Posted by 1337h4x
Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

Just plug in $0$ and $1$ into the polynomial.

5. Originally Posted by chiph588@
Just plug in $0$ and $1$ into the polynomial.
If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?

6. Originally Posted by 1337h4x
If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?
No when you plug in $0$ you get $1$.

Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.

All polynomials have roots. The question is determining those roots.

8. Originally Posted by wonderboy1953
All polynomials have roots. The question is determining those roots.
Over the complex numbers yes, but not an arbitrary field like $\mathbb{Z}/2\mathbb{Z}$.

9. Originally Posted by chiph588@
No when you plug in $0$ you get $1$.

Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.
Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?

10. Originally Posted by 1337h4x
Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?
Modulo 2 there are no roots what so ever.

Let $f(x)=x^4+x^2+1$.

$f(0)=1\not\equiv0\bmod{2}$
$f(1)=3\equiv1\not\equiv0\bmod{2}$

Thus there are no roots modulo $2$.

11. Originally Posted by chiph588@
Modulo 2 there are no roots what so ever.

Let $f(x)=x^4+x^2+1$.

$f(0)=1\not\equiv0\bmod{2}$
$f(1)=3\equiv1\not\equiv0\bmod{2}$

Thus there are no roots modulo $2$.
So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?

12. Originally Posted by 1337h4x
So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?
Read your original question. I showed you a polynomial f(x) that can be factored but there is no g such that $f(g)\equiv0 \bmod{2}$.

13. Ah, I see what you're saying. Thanks for making the connection. I lost track of what I was originally asking apparently!