## Polynomial Roots

#### Problem

Let $P(x) = x^n + c_{n-1}x^{n-1} + ... + c_{2}x^2 + c_{1}x + c_0$, where each of the coefficients, $c_{k}$, are integer. Prove that the roots of the equation $P(x) = 0$ will either be irrational or integer.

#### Solution

It is clear that irrational roots exist. For example, $P(x) = x^2 - 2 = 0 \Rightarrow x = \pm \sqrt{2}$.

So let us suppose that $x$ is rational and can be written as the ratio of two integers, $\dfrac{a}{b}$, such that $GCD(a,b) = 1$. By letting $x = \dfrac{a}{b}$ we get:

$\dfrac{a^n}{b^n} + c_{n-1}\dfrac{a^{n-1}}{b^{n-1}} + ... + c_{2}\dfrac{a^2}{b^2} + c_{1}\dfrac{a}{b} + c_0 = 0$

Multiplying through by $b^n$ gives:

$a^n + c_{n-1}a^{n-1}b + ... + c_{2}a^{2}b^{n-2} + c_{1}ab^{n-1} + c_{0}b^n = 0$

It can be seen that all terms are integer.

If we divide through by $b$ then the right hand side will be integer, but the left side can only be integer if $b \mid a^n$. As $a$ and $b$ are relatively prime, $b$ must equal 1, which means that if $x$ is rational then it must be integer.

Hence the roots of the equation $P(x) = 0$ will either be irrational or integer. Q.E.D.

Problem ID: 373 (07 Aug 2010)     Difficulty: 3 Star

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