an example of a ploynomial which factors, but has no roots

**beta12** · September 7th 2006, 02:52 AM

Give an example of a polynomial f(x) with integer coefficient which factors (poly mod n) but which has no roots, i.e. for which there are no integers x such that f(x) ≡ 0 (mod n).

I can't find it!

Can U help me? Thank you very much!

**ThePerfectHacker** · September 7th 2006, 06:49 AM

Originally Posted by beta12

Give an example of a polynomial f(x) with integer coefficient which factors (poly mod n) but which has no roots, i.e. for which there are no integers x such that f(x) ≡ 0 (mod n).

I can't find it!

Can U help me? Thank you very much!

$(x^2+1)(x^2+1)\equiv 0(\mbox{mod }p)$
For thus values of $p$ (prime) such that, the Legendre Symbol, $(-1/p)=-1$ , i.e. that is, $p=4k+3$
---
The reason why I used a prime is because $\mathbb{Z}_p[x]$ forms an integral domain and thus if two polynomials $a,b$ are such that $ab=0$ upon the evaluation homomorphism then $a=0\mbox{ or }b=0$ upon evaluation homomorphism.

**topsquark** · September 7th 2006, 10:12 AM

Originally Posted by ThePerfectHacker

$(x^2+1)(x^2+1)\equiv 0(\mbox{mod }p)$
For thus values of $p$ (prime) such that, the Legendre Symbol, $(-1/p)=-1$ , i.e. that is, $p=4k+3$
---
The reason why I used a prime is because $\mathbb{Z}_p[x]$ forms an integral domain and thus if two polynomials $a,b$ are such that $ab=0$ upon the evaluation homomorphism then $a=0\mbox{ or }b=0$ upon evaluation homomorphism.

I get it now! Good explanation. Now if only I could give you a rep point...

-Dan

**ThePerfectHacker** · September 7th 2006, 04:22 PM

Originally Posted by topsquark

Now if only I could give you a rep point...

Easy, spread some negative around then you can give me a postive point

**beta12** · September 8th 2006, 02:15 PM

Hi perfecthacker,

Thks for your reply. So far I have learnt up to polynomial congruence. I haven't learnt Legendre Symbol yet.

I have tried but still can't confirm your example is a polynomial which factors, but has no roots with my current polynomial congruence knowledge.

Could you please explain your example to me with the concept of polynomail congruence? Thank you very much.

**ThePerfectHacker** · September 8th 2006, 02:21 PM

Originally Posted by beta12

Hi perfecthacker,

Thks for your reply. So far I have learnt up to polynomial congruence. I haven't learnt Legendre Symbol yet.

I have tried but still can't confirm your example is a polynomial which factors, but has no roots with my current polynomial congruence knowledge.

Could you please explain your example to me with the concept of polynomail congruence? Thank you very much.

The Ploynomial,
$(x^2+1)(x^2+1)$ by definition is multiplied as,
$x^4+2x^2+1$
Thus, the polynomial,
$x^4+2x^2+1\equiv 0 (\mbox{mod }p)$
Is reducible (meaning it factors) but is has no zeros. Because, it is necessary and sufficient that,
$x^2+1\equiv 0(\mbox{mod }p)$
Equivalenty,
$x^2\equiv -1(\mbox{mod }p)$
Note, that we are claiming that there exists an $x$ such that is congruent to -1 modulo p. That is the same thing as saying "-1 is a quadradic residue of p" (definition). It can be shown without the Legendre Symbol that all the primes such that have -1 as a quadradic residue are 4k+1. In order, such that -1 not be a quadradic resides we require that p have form 4k+3 (because primes cannot have 4k,4k+2).

**beta12** · September 8th 2006, 02:38 PM

Thank you very much . I got it.

**ThePerfectHacker** · September 8th 2006, 02:42 PM

Originally Posted by ThePerfectHacker

It can be shown without the Legendre Symbol that all the primes such that have -1 as a quadradic residue are 4k+1.

Theorem:The congruence $x^2+1\equiv 0(\mbox{mod } p)$ where $p$ is an odd prime has a solution if and only if $p=4k+1$ .

Proof: Let $a$ solve the congruence. Thus,
$a^2\equiv -1(\mbox{mod } p)$
Fermat's Little Theorem,
$1\equiv a^{p-1}\equiv (a^2)^{(p-1)/2}\equiv (-1)^{(p-1)/2}(\mbox{mod } p)$
If $p=4k+3$ then the fact,
$(-1)^{(p-1)/2}=1$ would not hold.
That explain the "only if" part of the theorem.

Note that,
$(p-1)!=1\cdot 2\cdot ...\cdot \frac{p-1}{2}\cdot \frac{p+1}{2}\cdot ...\cdot (p-2)(p-1)$
Also note that,
$p-1\equiv -1$
$p-2\equiv -2$
$p-3\equiv -3$
...
$\frac{p+1}{2}\equiv -\frac{p-1}{2}$
Thus, the factorial above can be expressed as,
$(p-1)!\equiv 1(-1)2(-2)3(-3)\cdot ... \cdot \frac{p-1}{2} \cdot \left( -\frac{p-1}{2} \right)$
Thus,
$(p-1)!\equiv (-1)^{(p-1)/2} \left( 1\cdot 2\cdot 3\cdot...\cdot \frac{p-1}{2} \right)^2$
Wilson's Theorem,
$-1\equiv (-1)^{(p-1)/2} \left[ \left( \frac{p-1}{2} \right)!\right]^2(\mbox{mod } p)$ .
If,
$p=4k+1$
Then we have,
$\left[ \left( \frac{p-1}{2} \right)! \right]^2\equiv -1(\mbox{mod } p)$
Which is a solution to the congruence.

**beta12** · September 9th 2006, 01:08 PM

hi perfecthacker,

Why p-1 =-1?
from your proof
p-1 =-1
p-2=-2
:
:
etc. ?

I don't get this point. Could you please explain to me. Thank you very much!

**ThePerfectHacker** · September 9th 2006, 04:41 PM

Originally Posted by beta12

hi perfecthacker,

Hello

Originally Posted by beta12

Why p-1 =-1?

It is not equal, it is congruent to. But I did not place the modulo sign because I assumed you understood I was working modulo 'p'.

Originally Posted by beta12

from your proof

Not my proof.

Originally Posted by beta12

p-1 =-1
p-2=-2
:
:
etc. ?

What you end with is a system of congrunces. You multiply then out using the theorem that you can multiplly congruecnes under the same modulo.

**beta12** · September 11th 2006, 05:50 AM

Hi perfecthacker,

Thank you very much. I got it now.

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