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  "body": "\n## Proof\n\n1. Let $f = a_n x^n + \\dots + a_1 x + a_0 \\in \\Z[x]$ with $n \\ge 1$ and $a_n \\ne 0$. Let $r = p/q$ be a [[root|00EHX6FWC0]] of $f$, with $q \u003e 0$ and $\\gcd(p, q) = 1$.\n2. From $f(p/q) = 0$, multiply by $q^n$. This gives $a_n p^n + a_{n-1} p^{n-1} q + \\dots + a_1 p q^{n-1} + a_0 q^n = 0$.\n3. We show that $p^n$ and $q$ are [[coprime|0GFKP8F8AG]]. Suppose a [[prime|0SN0MAEKYX]] $\\ell$ divides both. Since $\\ell \\mid p^n = p \\cdots p$, [[Euclid's lemma|051VYKA1EX]] applied $n - 1$ times gives $\\ell \\mid p$.\n4. Then $\\ell$ divides $p$ and $q$, which contradicts $\\gcd(p, q) = 1$. So $\\gcd(p^n, q) = 1$. In the same way, $\\gcd(q^n, p) = 1$.\n5. By Step 2, $a_n p^n = -q \\left( a_{n-1} p^{n-1} + a_{n-2} p^{n-2} q + \\dots + a_0 q^{n-1} \\right)$. So $q$ divides $a_n p^n$.\n6. By Steps 4 and 5 and the [[generalized Euclid's lemma|03BAREW1Z5]], $q$ divides $a_n$.\n7. By Step 2, $a_0 q^n = -p \\left( a_n p^{n-1} + a_{n-1} p^{n-2} q + \\dots + a_1 q^{n-1} \\right)$. So $p$ divides $a_0 q^n$.\n8. By Steps 4 and 7 and the generalized Euclid's lemma, $p$ divides $a_0$.\n9. So $p$ divides $a_0$ and $q$ divides $a_n$.\n\n## Idea\n\nClearing denominators turns $f(p/q) = 0$ into an equation of integers. All terms but $a_0 q^n$ contain the factor $p$, and all terms but $a_n p^n$ contain the factor $q$. Since $p$ and $q$ share no prime factor, the divisibilities pass to $a_0$ and $a_n$.\n",
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