Matematické Fórum

stuart clark · 08. 12. 2014 04:45

A polynomial $f(x)$ has integer coefficients such that $f(0)$ and $f(1)$ are both odd numbers

Then How can we prove that $f(x)$ has no integer solutions

Brano · 08. 12. 2014 15:42 — Editoval Brano (08. 12. 2014 15:50)

Step 1: prove the following statement: Let $f$ be a polynomial with integer coefficients and let $x,k$ be integers, then $(x-k)|[f(x)-f(k)]$

Step 2: in your case let $x$ be an integer root, then $x|f(0)$ therefore $x$ is odd and $(x-1)|f(1)$ i.e. $f(1)$ is even, a contradiction.

Xellos · 13. 01. 2015 14:46

Look at parities. If a root $x$ is even, then $f(x)$ has the same parity as $f(0)$ , i.e. odd; if $x$ is odd, it has the same parity as $f(1)$ . Both are odd, so neither can be 0 and no such root exists.

Matematické Fórum

#1 08. 12. 2014 04:45

polynomial equation

#2 08. 12. 2014 15:42 — Editoval Brano (08. 12. 2014 15:50)

Re: polynomial equation

#3 13. 01. 2015 14:46

Re: polynomial equation

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