MHB Proving $f(x)=0$ Has No Integer Solution with Integer Coefficients

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A polynomial \(f(x)\) with integer coefficients has \(f(0)\) and \(f(1)\) as odd numbers, indicating that the constant term is odd and there is an even number of odd coefficients among the non-constant terms. This structure ensures that for any integer \(x\), \(f(x)\) remains odd. Consequently, since odd numbers cannot equal zero, \(f(x) = 0\) cannot have any integer solutions. Therefore, it is proven that \(f(x) = 0\) has no integer solutions.
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A polynomial $f(x)$ has Integer Coefficients such that $f(0)$ and $f(1)$ are both odd numbers. prove that $f(x) = 0$ has no Integer solution
 
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jacks said:
A polynomial $f(x)$ has Integer Coefficients such that $f(0)$ and $f(1)$ are both odd numbers. prove that $f(x) = 0$ has no Integer solution


There are details you may need to fill in yourself but:

\(f(0)\) odd implies that the constant term is odd

then \(f(1)\) odd implies that there are an even number of odd coeficients of the non constant terms.

So if \(x \in \mathbb{Z}\) then \(f(x)\) is odd and so cannot be a root of \(f(x)\) CB
 

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