Monic $f(x)\in\Bbb Z[x]$ has no rational root if $f(0)\ \&\ f(1)$ odd [Parity Root Test, Modular Root Test] A polynomial problem from my old algebra textbook:

$f(x)\in\Bbb Z[x]$ with leading coefficient $1$, $\deg f(x)\ge 1$, and both $f(0)$ and $f(1)$ are odd numbers, prove:
  $f(x)$ has no root within $\Bbb Q$

Eisenstein's criterion seems to be of little help here because we know virtually nothing about the coefficients. So I tried some other approach.
Let
$$f(x)=a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+x^n\in \Bbb{Z}[x]$$
What we have known so far is
$$f(0)=a_0\text{ is an odd number}$$
and
$$f(1)=a_0+a_1+a_2+\cdots+a_{n-1}+1 \text{ is an odd number}$$
from which we can also conclude that
$$f(1)-f(0)=a_1+a_2+\cdots+a_{n-1}+1\text{ is an even number}$$
To be honest I don't know how to proceed now. It isn't until just now that I realized "irreduicible" is not equivalent to "has no roots".. So all my previous attempt is fundamentally wrong..
Can anybody help or drop a hint? Best regards!
 A: Suppose that $f$ has a rational root $r$, then $r$ must be an integer by rational root theorem. Now observe that we have
$$
r-0|f(r)-f(0)=-f(0)\implies r\text{ is odd}
$$
And also
$$
r-1|f(r)-f(1)=-f(1)\implies r\text{ is even}
$$
Which contradicts...
A: Hint $\ $ By RRT = Rational Root Test and lead coef $= 1$, any rational root is an integer. But the Parity Root Test implies there are no integer roots (we can avoid use of RRT as here).
Remark $ $ We need only that the lead coef is $\rm\color{#c00}{odd}$, since then RRT implies that a lowest-terms rational root $\,a/b\,$ has $\rm\color{#c00}{odd}$ denominator $\,b\ $ so $\ {\rm mod}\ 2\!:\,\ b\equiv 1\,\Rightarrow\, a/b \equiv a\ $ is a root.
Modular Root Test $ $ The same method works for  larger moduli, i.e. if $\,f(x)\in\Bbb Z[x]\,$ is a polynomial with integer coefficients then any integer root $k$ persists $\!\bmod n,\,$ i.e. $\,f(k)=0\,\Rightarrow\, f(k)\equiv 0\pmod{\!n}\,$ by the Polynomial Congruence Rule. Thus if $\,f\,$ has no roots $\bmod n\,$ then $\,f\,$ has no integer roots. There are only a finite number of possible roots to test $\!\bmod n,\,$ viz $\,k\equiv 0,1,\ldots n-1,\,$ an effective (brute force) root test by modular case analysis.
The argument generalizes to  rings enjoying a sense of parity (or rings with an image of $\,\Bbb Z/n)$.

Update $\ $ OP is not familiar with the above modular arithmetic so we rewrite it in simpler language. The key idea is that the hypotheses imply that $\,f(n)\,$ is odd for all integers $\,n,\,$ so $\,f(n)\,$ never takes the even value $\,0.\,$ For proof, we consider separately the case $\,n\,$ is odd and $\,n\,$ is even.
Suppose $\,n = 1+2k\,$ is odd. Then by induction $\,n^i\,$ is odd for all natural $\,i\,$ since a product of odds is odd. Therefore we can write $\,n^i = 1 + 2k_i\,$ for integers $\,k_i.\,$ Substituting this into  $\,f(n) = \sum c_i n^i\,$ yields
$$\begin{align} f(n) &=\ c_0 + c_1(\color{#c00}1+\color{#0a0}{2k_1}) + c_2(\color{#c00}1+\color{#0a0}{2k_2})+\cdots+ c_d(\color{#c00}1+\color{#0a0}{2k_d})\\ 
&=\ \color{#c00}{c_0+c_1+c_2+\cdots + c_d} + \color{#0a0}{2(c_1k_1 + c_2 k_2 + \cdots + c_d k_d)}\\
&=\ \color{#c00}{f(1)} +\rm\color{#0a0}{even}\\  &=\ \rm odd+even\\  &=\rm\ odd\end{align}$$
The even case $\,n = 2k\,$ is similar but simpler: $\,f(2k) = f(0) + $ even = odd + even = odd. The details of this case are left to the reader.
