TITLE: Prove : The polynomial has no integral roots.
QUESTION [6 upvotes]: Q. Prove that a polynomial $f(x)$,with integer coefficients has no integral roots if $f(0)$ and $f(1)$ are both odd integers.
My attempt:
Let
$$f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$$
now $f(0)=a_0$ which is an odd integer.
and $f(1)=(a_0+a_1+a_2+\dots+a_n$) an odd integer. 
Now, what is the strategy I need to imply to prove that $f(x)$ can't have integral roots.

REPLY [2 votes]: The nonzero powers of an even number are even, the powers of an odd number are odd.
Then the value of $f$ for an even argument is the sum of $a_0$ and even terms so that
$$f(e)=a_0+a_1e'+a_2e''+\dots+a_ne^{(n)}$$ has the parity of $a_0=f(0)$.
And the value of $f$ for an odd argument is the sum of all coefficients times an odd factor so that
$$f(o)=a_0+a_1(e'+1)+a_2(e''+1)+\dots+a_n(e^{(n)}+1)$$ has the parity of $a_0+a_1+a_2+\cdots a_n=f(1)$.
Conclusion: for any integer, the function value is odd, thus nonzero.