1. ## 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.

2. ## Re: Non-existence

If $p(x)q(x)=x^5+ 2x+1$, then either
Case 1
$p$ is of degree $1$ and $q$ is of degree $4$,
or
Case 2
$p$ is of degree $2$ and $q$ is of degree $3$

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

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

Contradiction has occurred, so Case 1 is impossible
Case 2 is proven similarly

3. ## Re: Non-existence

I just wish to know if the following reasoning is correct?

Working few steps as given by I-Think above:
Originally Posted by I-Think
If $p(x)q(x)=x^5+ 2x+1$, then either
Case 1
$p$ is of degree $1$ and $q$ is of degree $4$,
or
Case 2
$p$ is of degree $2$ and $q$ is of degree $3$

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

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

4. ## Re: Non-existence

Yes, it is a very correct and much shorter and neater way of answering the question, for case 1.

5. ## 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