# Rational roots existence proof

Problem statement

We want to show that the polynomial $x^p-p\,2^p\,x+p^2\in\mathbb Z[x]$, $p$ prime, has no rational root.

My approach

We separate the proof in two steps: $p>2$ and $p=2$ ($p<0$ is not possible because coefficients would not be in $\mathbb Z$).

If $p=2$, the polynomial is $x^2-8\,x+4$, and its roots are: $2\,(2\pm\sqrt{3})$, but they are not rational because $\sqrt{3}\notin\mathbb Q$.

For $p>2$ we use a RA argument. We assume that exists $\frac{a}{b}$, with $a,\,b\in\mathbb Z$ coprime, that satisfies:

$$\frac{a^p}{b^p}-p\,2^p\,\frac{a}{b}+p^2=0,$$

if they are not coprime we can simplify until $\mbox{gcd}(a,\,b) = 1$. Then, multiplying by $b^p$, we have:

$$a^p-p\,2^p\,a\,b^{p-1}+p^2\,b^p=0.$$

On the one hand, if we isolate $a^p$ and take $b$ as common factor:

$$b\,\left(-p\,2^p\,a\,b^{p-2}+p^2\,b^{p-1}\right)=-a^p,$$

so $b\mid -a^p$. As $\mbox{gcd}(a,\,b) = 1$, $b$ divides $a$ and they are coprimes, hence $b=1$. On the other hand, if we isolate the independent term and take $a$ as common factor:

$$a\,\left(a^{p-1}-p\,2^p\,b^{p-1}\right)=-p^2\,b^p.$$

Thus, $a\mid -p^2\,b^p$. As $\mbox{gcd}(a,\,b) = 1$ we have that $a\mid p^2$. Since $p$ is prime the only possible rational roots are: $\pm 1$, $\pm p$ or $\pm p^2$.

Let's see that they are not roots:

• $\pm1$ is not a root.

$$\pm1\mp p\,2^p+p^2=0 \iff p\,(2^p\pm p) = 0$$

Which is not possible because $(2^p\pm p)$ would be the inverse of $p$ but $(2^p\pm p)\in\mathbb Z$.

Difficulty

I get stuck with the roots $\pm p$ and $\pm p^2$. I don't know how to see that they are not real roots.

EDIT: Another approach

May be, if we take $p$ as common factor instead of $b$,

$$p\,\left(-2^p\,a\,b^{p-1}+p\,b^{p}\right)=-a^p,$$

then $p\mid -a^p$ but, as $p$ is prime, $\mbox{gcd}(p,\,a)=1$, hence $p=1$ and we get a contradiction.

Would this be correct?

-
May be there is an easiest way... – synack Oct 28 '12 at 10:29
What about the rational root theorem ? if there is a root then it is plus minus of $1,p,p^2$ – Belgi Oct 28 '12 at 10:41
Yes, it is what I'm using (see the Rational Theorem proof). The problem is that I want to see that $\pm p$ and $\pm p^2$ are not roots. – synack Oct 28 '12 at 10:57

If $p\geq 3$ were a root, then $p^p-p^22^p+p^2=0$, so $2^p-1=p^{p-2}$ would be a multiple of $p$, but Fermat theorem says that $2^p\equiv 2$ (mod $p$), so we get a contradiction.

If $p^2$ were a root, then $p^{2p}-p^32^p+p^2=0$, so $p2^p-1=p^{2p-2}$ would be a multiple of $p$, but it is clear that $p2^p-1$ is not a multiple of $p$.

Observe that, in principle, $-p$ and $-p^2$ might also be roots of the polynomial, but a similar reasoning can be made to discard them.

-
$- p$ and $- p^2$ cannot be, because I assume that $p$ is positive. Otherwise the polynomial would have not integer coefficients. Right? – synack Oct 28 '12 at 10:48
I don't agree with that, because $-p$ and $-p^2$ are also divisors of $p^2$ (it does not matter the sign of $p$), so you cannot discard them. – Manzano Oct 28 '12 at 10:50
Ok, thanks. What about the other approach that I edited? – synack Oct 28 '12 at 10:56
I think that from $p|-a^p$ you only get that $p|a$, because you cannot be sure that gcd$(a,p)=1$, – Manzano Oct 28 '12 at 11:08
Ok, this way you could know that $a\neq \pm 1$, because $p \mid a$. Right? – synack Oct 28 '12 at 11:13

Take $\nu_2(q)$ as the maximum power of $2$ that divides the numerator of the rational number $q$. If $q$ is a root of your polynomial and $p$ is odd,

$$p\cdot\nu_2(q) = p+\nu_2(q)$$

must hold, that is clearly impossible.

-
I don't understand your answer... Could you explain it a little bit deeper? – synack Oct 29 '12 at 8:02