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A polynomial equation with natural coefficients can have natural roots ?

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You mean natural number coefficients and roots? –  Alex Becker Apr 28 '12 at 19:16
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If the natural numbers contain $0$, then yes: $x=0$ is an example.

Otherwise, no, because for all $x \in \mathbb N $, $a_n x^n+a_{n-1}x^{n-1}+ \dots +a_1 x +a_0 \in \mathbb N \not\ni 0$.

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Suppose $a_n x^n+a_{n-1}x^{n-1}+ \dots +a_1 x +a_0$ is a polynomial with $a_j$ positive integers (natural numbers). Suppose that $x_1, x_2, \dots, x_n$ are the roots; all positive integers. Then $$ x_1+x_2+ \dots+x_n =-\frac{a_{n-1}}{a_n} $$ a contradiction.

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Vieta's formulae are always very handy... –  J. M. Apr 28 '12 at 19:50
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No.

Suppose a polynomial has at least one natural root. Then it can be written as $$(x - r)(a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1x + a_0)$$ which equals $$a_{n-1}x^n + (a_{n-2} - ra_{n-1})x^{n-1} + (a_{n-3} - ra_{n-2})x^{n-2} + \dots + (a_0 - ra_1)x - ra_0.$$

In order for the $x^n$ term to be positive, $a_{n-1}$ must be positive. But then in order for the $x^{n-1}$ term to be positive, the $a_{n-2}$ term must be positive (since $r$ is positive and we must outweigh the contribution of $-ra_{n-1}$). If we keep going, we get that the $a_0$ term must be positive. However, the very last term is $-ra_0$, which is then negative. This means the polynomial cannot have all natural coefficients.

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