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chiph588@ · #9 February 9th, 2009, 09:02 PM

Proof by contradiction: Assume $f(x)$ is a polynomial with integer roots.

First note that $f(0)$ is the product of all the roots and $f(1)$ is the sum of coefficients.

Let $f(x) = (x-r_1) \cdots (x-r_n)$ , where $r_1, \cdots , r_n$ are all the roots to $f(x)$ .

So $f(0) = \prod_{i = 1}^n (-r_i)$ and since $f(0)$ is odd, this implies each $r_i$ is odd too.

But $f(1) = \prod_{i = 1}^n (1-r_i)$ and since $f(1)$ is odd, this implies that each $1-r_i$ is odd which implies each $r_i$ is even.

Hence a contradiction and therefore there are no integer solutions.