This question arises in STEP 2011 Paper III, question 2. The paper can be found here.

The first part of the question requires us to prove the result that if the polynomial $$x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{0}$$ where each of the $a_{n}$ are integers, has a rational roots if and only if that root is an integer. It does not give a name for this result.
EDIT: user69810 has pointed out that this is in fact the rational root theorem.

We are then to use this result to prove that the polynomial $$x^{n}-5x+7=0$$ has no rational solutions for $n\ge 2$. My argument was the following:
If there exists a rational root, then it must be an integer.
If there exists an integer root, then $$x^{n}=5x-7$$ for some integers $x$ and $n\ge 2$. Then the LHS is divisible by $x$, meaning that $7$ must be divisible by $x$. Therefore $x \in \{-7,-1,1,7\}$. Checking each of these shows that there is no rational root.

Is this solution correct? It isn't the one in the solutions, but if it's right then I think it's more elegant than theirs.

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    $\begingroup$ The result is called the rational root theorem. $\endgroup$ – John Douma Jun 12 '13 at 19:54
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    $\begingroup$ Or the Gauss lemma, which immediately gives that $x-a|x^n+\ldots +b\in\mathbb Z[x]$ implies $a|b$. $\endgroup$ – Hagen von Eitzen Jun 12 '13 at 19:58
  • $\begingroup$ Looks fine, but its probably more straightforward to say $x^n-5x=-7 \implies x|x^n-5x \implies x|7$. $\endgroup$ – Gamma Function Jun 12 '13 at 21:09

You have identified the possible rational roots correctly and shown that none of them work. It is a fine application of the rational root theorem.

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  • $\begingroup$ Excellent, thank you! $\endgroup$ – preferred_anon Jun 12 '13 at 20:03

Yes, it's correct. Simpler: Rational Root Test $\,\Rightarrow\,x\in\Bbb Z\,\Rightarrow\, 7 = 5x\!-\!x^n\,$ is even, contradiction.

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  • $\begingroup$ I like this! +1 $\endgroup$ – preferred_anon Jun 13 '13 at 12:33

I think it's not correct, because you are assuming that if $x|5x-7$ then $x|7$. This is not true in general.Think for example to $2+4$. Of course $6|2+4$ but $6$ doesn't divide $2$ nor $4$. What you can say is that $x(x^{n-1}-5)=7$, but $7$ is prime, then you must have $x=1$ and $x^{n-1}-5=7$ or $x=7$ and $x^{n-1}-5=1$. But both are absurd because $x=1$ and $x=7$ doesn't solve the equation. As you can see the result is quite the same, but the procedure is quite different for what you assumed at the beginning. Just a remark, what you called Eisenstein criterion is Gauss Lemma, Eisenstein criterion is about irriducibility in $\mathbb{Q}[x]$.

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  • $\begingroup$ since $x|5x$, concluding $x|7$ is absolutely correct. $\endgroup$ – pascalhein Jun 12 '13 at 20:23
  • $\begingroup$ No, it's true: $\ x\mid5x,\,5x\!-\!7\,\Rightarrow\,x\mid 5x\!-\!(5x\!-\!7) = 7.\ \ $ $\endgroup$ – Key Ideas Jun 12 '13 at 21:24

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