In the following equation:
$r^3-7r+1=0$
why $r$ cannot be rational?
So far my considerations are the following.
Suppose $r$ is rational. Thus it can be presented in the form $r=\frac{p}{q}$, where $p$ and $q$ are integers and co-prime, $q\neq0$.
$\frac{p^3}{q^3}-7\frac pq+1=0 | \times q^3 \iff p^3 - 7pq^2 + q^3 = 0 \iff p^3 + q^3 = 7pq^2$.
But at this level I got stuck in finding the contradiction. Any hint is appreciated. Thanks!
Let us rearrange your equation in a different way.
First, $p^3=7pq^2-q^3=q(7pq-q^2)$. Now, since the right hand side is a multiple of $q$, we have $q|p^3$, but also $p$ and $q$ are (by assumption) relatively prime. Therefore, if there were a solution, $q=\pm 1$.
Now apply similar reasoning to $q^3=7pq^2-p^3=p(7q^2-p^2)$.
This gives you a short list of candidates for $p$ and $q$, hence for $r$. Do any of them work?
It is already naturally imposed that $q \neq 0$. Note that $p \neq 0$; if otherwise then $0 = 1$. Note that $p^{3}+q^{3} = 7pq^{2}$ implies that $q$ divides $p^{3}$. If $q = \pm 1$, then $p^{3} \pm 1 = 7p$, so $p$ divides $\pm 1$, hence $p=\pm 1$, but then $0, 2= \pm 7$, a contradiction. So suppose $p, q \neq \pm 1$. Still it holds that $q$ divides $p^{3}$, which is again impossible as $p,q$ were coprime.
The polynomial $r^3-7r+1$ is irreducible over $\mathbb{F}_2$, hence it is irreducible over $\mathbb{Q}$.
In particular it does not have any rational root.
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$\ds{p^{3} + q^{3} = 7pq^{2}\,,\qquad\mbox{LHS} = p^{3} + q^{3}\,,\quad\mbox{RHS} = 7pq^{2}}$.
\begin{align} \pars{p\ odd,q\ odd} & \implies \pars{\mbox{LHS},\mbox{RHS}} = \pars{even,odd} \\[1mm] \pars{p\ odd,q\ even} & \implies \pars{\mbox{LHS},\mbox{RHS}} = \pars{odd,even} \\[1mm] \pars{p\ even,q\ odd} & \implies \pars{\mbox{LHS},\mbox{RHS}} = \pars{odd,even} \\[1mm]\require{cancel} \cancel{\pars{p\ even,q\ even}} & \implies \cancel{\pars{\mbox{LHS},\mbox{RHS}}} = \cancel{\pars{even,even}}:\ p\ \mbox{and}\ q\ \mbox{have common factors.} \end{align}
(p even, q even) creates a contradiction. What about the other 3 cases?
– Dmitri Pavlutin
Jun 04 '17 at 15:14
Hint (another way):
All rational solutions of the equation must satisfy: $q$ divides the leading coefficient (in this case $1$) and $p$ divides the constant term (in this case $1$).
$p\over q$ must be in lowest terms.
You have a handful of cases to check.
Direct followed from your attempt.
As $$p^3+q^3=7pq^2$$ $$q^3=p(7q^2-p^2)$$ Seen that $7q^2-p^2\in\mathbb Z$, so $p$ must be a factor of $q^3$, implies that $p$ and $q$ have some factors are same, which already contradicts the assumption that $p$ and $q$ are co-prime if $p\neq\pm 1$.
It is easy to check that $p$ is not $1$ by $vice\ versa$ on $q$. (If so then $q=\pm 1\Rightarrow r=\pm 1$, also a contradiction.)
