Integer coefficients of cubic equation imply integer roots

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Problem: Let $a,b,c$ be three integers for which the sum$ \frac{ab}{c}+ \frac{ac}{b}+ \frac{bc}{a}$is integer. Prove that each of the three numbers $ \frac{ab}{c}, \quad \frac{ac}{b},\quad \frac{bc}{a} $is integer.

(Proposed by Gerhard J. Woeginger)

The solution is as follows: Set $u := ab/c, v := ac/b$ and $w := bc/a$. By assumption, $u + v + w$ is an integer. It is easily seen that $uv + uw + vw = a^2 + b^2 + c^2$ and $uvw = abc$ are integers, too. According to Vieta’s formulæ, the rational numbers $u, v, w$ are the roots of a cubic polynomial$x^3 + px^2 + qx + r$ with integer coefficients. As the leading coefficient is 1, these roots are integers.

Question: I haven't found anything supporting this, but why does a cubic equation with integer coefficients and a leading coefficient of 1 have integer roots? Why does the cubic equation $x^3+5x^2-20x+5=0$ not have integer roots then?

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1 Answer

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It follows from the rational root theorem that every rational root of every monic polynomial (of any degree) with integer coefficients must be integer.

Besides, the polynomial $x^3+5x^2-20x+5$ has no rational roots.

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