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Let $P(x)$ be a polynomial with integer coefficients. In what conditions that $P(x)$ doesn't have a rational root?

From https://en.wikipedia.org/wiki/Rational_root_theorem, if $P(x)=a_nx^n+\cdots +a_0$ with integer coefficients, where $a_0\ne 0$, $a_n\ne 0$, then the only conceivable rational roots of $P(x)$ are of the form $\dfrac{a}{b}$, where $a|a_0$, $b|a_n$. However there are some cases that for $a|a_0$, $b|a_n$, $\dfrac{a}{b}$ migh not be a root of $P(x)$.

Are there any general criterion so that $P(x)$ doesn't have a rational root?

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  • $\begingroup$ en.wikipedia.org/wiki/Rational_root_theorem $\endgroup$ Commented Oct 25, 2018 at 20:50
  • $\begingroup$ What are you hoping for here? The rational root theorem gives you a good way to test a given polynomial. Note: this has little to do with combinatorics or number theory. $\endgroup$
    – lulu
    Commented Oct 25, 2018 at 20:50
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    $\begingroup$ @lulu It has much to do with number theory. $\endgroup$ Commented Oct 25, 2018 at 20:55

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In general, if we have a polynomial $P(x)$ with integer coefficients, where $P(x)=a_0x^n+\cdots +a_n$, where $a_0\ne 0$, $a_n\ne 0$, then the only conceivable rational roots of $P(x)$ are of the form $\dfrac{a}{b}$, where $a$ is a divisor (possibly negative) of $a_n$ and $b$ is a positive divisor of $a_0$.

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    $\begingroup$ Thanks for your answer. However, I want to know when $P(x)$ doesn't have a rational root. If $a|a_n$ and $b|a_0$, $\frac{a}{b}$ might not be a root of $P(x)$ $\endgroup$
    – apple
    Commented Oct 25, 2018 at 21:00
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    $\begingroup$ I think the recurring answer to your question is rational root theorem, because it's an easy way to check if a polynomial has a rational root. Therefore it's an easy way to check if a polynomial does not have a rational root. Other than that, I don't know any of the top of my head and if I think of one I'll post it here $\endgroup$
    – NazimJ
    Commented Oct 26, 2018 at 1:21

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