# Polynomials which factors but have no roots?

• May 28th 2010, 05:53 AM
1337h4x
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!
• May 28th 2010, 08:36 AM
chiph588@
Quote:

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 $2$.
• May 28th 2010, 08:55 AM
1337h4x
Quote:

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?

• May 28th 2010, 09:07 AM
chiph588@
Quote:

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.
• May 28th 2010, 09:15 AM
1337h4x
Quote:

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?
• May 28th 2010, 09:23 AM
chiph588@
Quote:

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.
• May 28th 2010, 09:27 AM
wonderboy1953
All polynomials have roots. The question is determining those roots.
• May 28th 2010, 09:31 AM
chiph588@
Quote:

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 $\mathbb{Z}/2\mathbb{Z}$.
• May 28th 2010, 09:43 AM
1337h4x
Quote:

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?
• May 28th 2010, 10:01 AM
chiph588@
Quote:

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 $2$.
• May 28th 2010, 10:17 AM
1337h4x
Quote:

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 ?
• May 28th 2010, 10:27 AM
chiph588@
Quote:

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 $f(g)\equiv0 \bmod{2}$.
• May 28th 2010, 10:37 AM
1337h4x
Ah, I see what you're saying. Thanks for making the connection. I lost track of what I was originally asking apparently!