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Exercise 1826Algebra › PolynomialsOlympiadAdded 3 Nov 2026

A polynomial with integer coefficients

#integer coefficients#roots#Gauss lemma

Statement

Let P(X)=anXn+⋯+a0P(X)=a_nX^n+\cdots+a_0 be a polynomial with integer coefficients, an≠0a_n\neq0. Show that if p/qp/q is a rational root in lowest terms, then p∣a0p\mid a_0 and q∣anq\mid a_n.

Hint

Multiply P(p/q)=0P(p/q)=0 by qnq^n.

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Solution

Setup. P(p/q)=0P(p/q)=0 gives
an(p/q)n+an−1(p/q)n−1+⋯+a0=0.a_n(p/q)^n+a_{n-1}(p/q)^{n-1}+\cdots+a_0=0.
Multiply by qnq^n:
anpn+an−1pn−1q+⋯+a0qn=0.a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n=0.

p∣a0p\mid a_0. Modulo pp: a0qn≡0(modp)a_0q^n\equiv0\pmod p. Since gcd⁡(p,q)=1\gcd(p,q)=1, p∣a0p\mid a_0.

q∣anq\mid a_n. Modulo qq: anpn≡0(modq)a_np^n\equiv0\pmod q, so q∣anq\mid a_n.

Conclusion. The rational root test holds.