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Number theory Difficulty 6.2 National olympiad Prove it

2. If x2+ax+b=0x^{2}+a x+b=0 has an integer root x0≠0x_{0} \neq 0, then x0∣bx_{0} \mid b. Generally, if
xn+an−1xn−1+⋯+a0=0x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}=0

has an integer root x0≠0x_{0} \neq 0, then x0∣a0x_{0} \mid a_{0}.

Solution

2. a0=−x0(x0n−1+an−1x0n−2+⋯+a1)a_{0}=-x_{0}\left(x_{0}^{n-1}+a_{n-1} x_{0}^{n-2}+\cdots+a_{1}\right).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.