Root of x^n + c_(n-1) x^(n-1) + ... + c_1 x + c_0 is either an integer or an irrational where all c

Let \(r\) be the root of the polynomial \(x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0\), where \(c_0, c_1, \ldots, c_{n-1}\) are integers, and let \(r = a/b\), where \(a/b\) is a fraction in it's smallest form (where \(a\) and \(b\) are relatively prime integers), and where \(|b| \gt 1\).

If \(r\) is the root:

$$\begin{gather} (a/b)^n + c_{n-1}(a/b)^{n-1} + \cdots + c_1 (a/b) + c_0 = 0 \\ a^n + c_{n-1}a^{n-1}b + \cdots + c_1 ab^{n-1} + c_0 b^n = 0 \\ a^n = b * -(c_{n-1}a^{n-1} + \cdots + c_1 ab^{n-2} + c_0 b^{n-1}) \\ a^n = bm \end{gather}$$

We see that \(b|a^n\). Since \(|b| \gt 1\), then \(b\) has at least one prime divisor \(p\). If \(p|b\), then \(p|a^n\). All integers, including \(a\), has a unique prime factorization:

$$\begin{gather} a = p_1^{e_1}p_2^{e_2} \ldots p_n^{e_n} \\ a^n = p_1^{ne_1}p_2^{ne_2} \ldots p_n^{ne_n} \end{gather}$$

If \(p|a^n\), then \(p\) has to be one of \(p_1,p_2,\ldots,p_n\) which means \(p|a\). If \(p|b\) and \(p|a\), then \((a,b)\gt 1\), which is a contradiction. Therefore, \(r\) cannot be of the form \(a/b\) where \(|b| \gt 1\). Either \(r\) is irrational or an integer \((|b|=1)\).

Root Of xⁿ+c_n-1x^n-1+...+c₁x+c₀ Is Either An Integer Or An Irrational Where All c_i's are integers

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