Ivan Shishkin, Rye (1878)

Problems/Number theoryUnreviewed

Rational roots of a monic polynomial in Z[X]\Z[X]

by Uettechat·translated by visitor·
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Show that if rQr \in \mathbb{Q} is a root of a monic polynomial PZ[X]P \in \mathbb{Z}[X], then rZr \in \mathbb{Z}.

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Solution by UettechatFR

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On note a0,...,an1,ana_{0},...,a_{n-1}, a_{n} les coefficients de PP (donc an=1a_{n}=1)
Soient p,qZ×N,pgcd(p,q)=1, tel que r=pqQp,q \in \mathbb{Z} \times \mathbb{N^{*}}, pgcd(p,q)=1, \text{ tel que } r=\frac{p}{q} \in \mathbb{Q} soit racine de PP.

Il vient P(r)=a0+a1pq+...+an1(pq)n1+(pq)n=0P(r)=a_{0}+a_{1}\frac{p}{q}+...+a_{n-1}(\frac{p}{q})^{{n-1}}+(\frac{p}{q})^{{n}}=0
Donc pn=an1pn1q...a1pqn1a0qn=q(an1pn1...a1pqn2a0qn1)p^{n}=-a_{n-1}p^{n-1}q-...-a_{1}pq^{n-1}-a_{0}q^{n}=-q(a_{n-1}p^{n-1}-...-a_{1}pq^{n-2}-a_{0}q^{n-1})
Donc qpnq | p^{n}, or pgcd(p,q)=1pgcd(p,q)=1, donc qpq|p.
Il vient que q{1,1}q \in \{-1,1\}.
Donc r=pZr=p \in \mathbb{Z}

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