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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?

  • 0
    Yes, it is what I'm using (see the [Rational Theorem proof](http://en.wikipedia.org/wiki/Rational_root_theorem)). The problem is that I want to see that $\pm p$ and $\pm p^2$ are not roots.2012-10-28

2 Answers 2

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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.

  • 0
    Yes, that's true, it's what I did. Thank you @Manzano, I'll check your answer as correct.2012-10-29
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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.

  • 0
    I don't understand your answer... Could you explain it a little bit deeper?2012-10-29