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?