May 14, 2012 Thread starter #1 J jacks Well-known member Apr 5, 2012 226 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
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
May 14, 2012 #2 C CaptainBlack Well-known member Jan 26, 2012 890 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 Click to expand... 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
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 Click to expand... 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