# Non-existence

• Nov 16th 2011, 12:48 AM
malaygoel
Non-existence
Show that there do not exist polynomials p(x) and q(x) of degree greater than or equal to 1 with integer co-efficients such that
p(x)q(x)=x^5+ 2x+1.
• Nov 17th 2011, 07:17 AM
I-Think
Re: Non-existence
If $\displaystyle p(x)q(x)=x^5+ 2x+1$, then either
Case 1
$\displaystyle p$ is of degree $\displaystyle 1$ and $\displaystyle q$ is of degree $\displaystyle 4$,
or
Case 2
$\displaystyle p$ is of degree $\displaystyle 2$ and $\displaystyle q$ is of degree $\displaystyle 3$

I'll do case 1
$\displaystyle (ax+b)(cx^4+dx^3+ex^2+fx+g)=x^5+2x+1$
But the polynomials have integer coefficients, so $\displaystyle a=c=\pm{1}$ and $\displaystyle b=g=\pm{1}$
(I'm going to choose $\displaystyle +1$ for both to simplify things here)

$\displaystyle (x+1)(x^4+dx^3+ex^2+fx+1)=x^5+2x+1$
Expanding
$\displaystyle x^5+(d+1)x^4+(e+d)x^3+(e+f)x^2+(f+1)x+1=x^5+2x+1$
Comparing coefficients
$\displaystyle d+1=0$, so $\displaystyle d=-1$
$\displaystyle e+d=0$, so $\displaystyle e=1$
$\displaystyle e+f=0$, so $\displaystyle f=-1$
$\displaystyle f+1=2$, so $\displaystyle f=1$

Contradiction has occurred, so Case 1 is impossible
Case 2 is proven similarly
• Nov 17th 2011, 12:01 PM
malaygoel
Re: Non-existence
I just wish to know if the following reasoning is correct?

Working few steps as given by I-Think above:
Quote:

Originally Posted by I-Think
If $\displaystyle p(x)q(x)=x^5+ 2x+1$, then either
Case 1
$\displaystyle p$ is of degree $\displaystyle 1$ and $\displaystyle q$ is of degree $\displaystyle 4$,
or
Case 2
$\displaystyle p$ is of degree $\displaystyle 2$ and $\displaystyle q$ is of degree $\displaystyle 3$

I'll do case 1
$\displaystyle (ax+b)(cx^4+dx^3+ex^2+fx+g)=x^5+2x+1$
But the polynomials have integer coefficients, so $\displaystyle a=c=\pm{1}$ and $\displaystyle b=g=\pm{1}$
(I'm going to choose $\displaystyle +1$ for both to simplify things here)

$\displaystyle (x+1)(x^4+dx^3+ex^2+fx+1)=x^5+2x+1$

Now, since x=-1 is a root of LHS but not of RHS, LHS cannot be equal to RHS for any values of d,e and f. Is it correct to conclude like this?
• Nov 18th 2011, 08:04 AM
I-Think
Re: Non-existence
Yes, it is a very correct and much shorter and neater way of answering the question, for case 1.
• Nov 18th 2011, 10:08 AM
Also sprach Zarathustra
Re: Non-existence
Maybe another solution(only highlights):

x^5+ 2x+1=0 have only one real root (wolframalpha or calculus laws).

Suppose that m is root of x^5+ 2x+1, now, we need to prove that m is not rational.

Lets say that m=p/q , where p and q are integers.

m^5+ 2m+1=(p/q)^5+ 2(p/q)+1=0

==>

p^5+2p*q^4+q^5=0

Integer solution only when p=q=0

Contradiction! Therefore, m is irrational.