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This questions contains a comment that states that $\gcd(r,s)\ne 1$ (which leads to a contradiction) where our rational root is $r/s$.

Why would that be true?

Bernard
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John Cataldo
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  • See Wikipedia. – Jyrki Lahtonen May 15 '18 at 21:03
  • Write the equation as $-r^n=s(a_{n-1}r^{n-1}+a_{n-2}r^{n-2}s+...+a_0s^{n-1})$. Then a prime divisor of $s$ divides the righ-hand side, and therefore it must divide the left-hand side ($-r^n$), and therefore, it must divide $r$. This means that any prime divisor of $s$ also divides $r$. If $s\neq1$, then such prime divisors of $s$ exist. This would imply that $gcd(s,r)>1$. –  May 15 '18 at 21:17

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Multiplying by $s^n$ gives $$r^n + a_{n-1}r^{n-1}s + \dots+ a_1rs^{n-1} + a_0s^n = 0$$ and so $$r^n=-s(a_{n-1}r^{n-1}+\dots+a_1rs^{n-2}+a_0s^{n-1}).$$ Thus $s\mid r^n$, which implies any prime factor of $s$ is also a prime factor of $r$. So, we conclude $\gcd(s,r)\neq 1$ unless $s$ has no prime factors.

Eric Wofsey
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