Finding (guessing) roots

Task number: 3065

Let \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \] be a polynomial of degree \(n\) with integer coefficients (i.e. all \(a_i\) are integers). Assume that \(P(x)\) has a rational root \(c = \frac pq\), where \(\frac pq\) is an irreducible fraction. Prove that \(a_0\) is divisible by \(p\) and \(a_n\) is divisible by \(q\).

  • Hint

    Expand the equation \(P(p/q) = 0\).

  • Resolution

    If we expand \(P(p/q) = 0\), we obtain \[ a_n \frac{p^n}{q^n} + a_{n-1}\frac{p^{n-1}}{q^{n-1}} + \cdots + a_1 \frac pq + a_0 = 0 \] or \[ a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 pq^{n-1} + a_0q^n = 0. \] All terms \(a_i p^i q^{n-i}\) are divisible by \(p\) for \(i \geq 1\). So even the last term \(a_0 q^n\) is divisible by \(p\). But the values \(q^n\) are relatively prime to \(p\), because \(p/q\) is an irreducible fraction. So \(a_0\) is divisible by \(p\). Similarly we can show that \(a_n\) is divisible by \(q\).

Difficulty level: Easy task (using definitions and simple reasoning)
Proving or derivation task
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