Integer coefficient polynomial $p(x)$ has no integer roots of $\,p(0)$ and $p(1)$ are odd [Parity Root Test] This came up an a training piece for the Putnam Competition and also in Ireland and Rosen.
The question posed was basically:
Let $p(x)$ be a polynomial with integer coefficients satisfying that $p(0)$ and $p(1)$ are odd. Show that $p$ has no integer zeros.
I&R give an example:
$p(x) = x^2 - 117x + 31$  and show (no problem) that for any $n$ whether even or odd, $p(n)$ will be odd. And claim that this shows $p(n)$ will never be $0$.
I can see, e.g., that $x^2 + 2x + 1$ will be odd substituting an even $n$ and even for an odd $n$. 
But would appreciate help in understanding the underlying math and what is happening here.
Also, as a second part, can a general statement about the existence of an integer solution be made if $n$, even and odd, generates an even and an odd as in the last example.
I can see that if you look at these equations (mod $2$), you can distinguish whether there is an integer solution. And I would guess this is intimately connected with the question.
Thanks as always.
 A: Hint: What can you say about $f(x) \pmod{2}$?
A: Hint $\ $  If an integer coefficient polynomial has an integer root $\rm\,n,\,$ i.e.$\rm\ p(n) = 0,\ $ then $\rm\,n\,$ remains a root modulo $2$, i.e. $\rm\ p(n)\equiv 0\,\ (mod\ 2).\:$ So, contrapositively, if a polynomial has no roots modulo $2$ then it has no integer roots. This leads to the following simple 
Parity Root Test $\ $ A polynomial $\rm\:P(x)\:$ with integer coefficients 
has no integer roots if its constant coefficient and coefficient sum are both odd. 
Proof $\ $ The test  verifies that $\rm\  P(0) \equiv  1\equiv P(1)\ \ (mod\ 2)\:,\ $ i.e. 
that $\rm\:P(x)\:$ has no roots modulo $2$, hence no integer roots. $\ $ QED 
E.g. $\rm\:\ a\ X^2 + b\ X + c\ $  has no integer roots 
if  $\rm\:c\:$  is odd and $\rm\:a,\:b\:$  have equal parity $\rm\:a\equiv b\ (mod\ 2)$
The Parity Root Test generalizes to any ring with a sense of parity, e.g. the Gaussian integers $\rm\: a + b\,{\it i}\ $ for integers $\rm\:a,b.\:$ For much further discussion see this post and also these related posts.
A: If you plug an integer into a polynomial and this equals zero, then you can look at the entire computation mod your favorite $n$. In this case, try two. What does the equation look like then?
A: If $p(0)$ and $p(1)$ are both odd, then $p(x)$ cannot have any integer roots.  If it did, then there would be a solution to $p(x)\equiv0$ mod $2$.  But $p(0)\equiv p(1)\equiv1\not\equiv0$ mod $2$.
If you're not comfortable with modular arithmetic here, suppose $p(x)=(x-r)q(x)$ with an integer root $r$.  Then $-rq(0)=p(0)$ implies $r$ is odd, but $(1-r)q(1)=p(1)$ implies $1-r$ is odd, which implies $r$ is even.  Taken together, we have a contradiction.
A: If there is an integer solution to $f(x)=0$, say $a$, then, if $a$ is even, $f(a)\equiv f(0)\equiv 1\mod 2$ and if $a$ is odd, $f(a)\equiv f(1)\equiv 1\mod 2$. So, no integer solution is possible.
A: The mod $2$ thing is easiest. Here is another idea.
Note that the polynomial expression $f(x)-f(y)$ is divisible by $x-y$ because $x^r-y^r=(x-y)(x^{r-1}+x^{r-2}y+\dots + xy^{r-2}+y^{r-1})$ and the polynomial is built up from such expressions.
So $f(n)-f(0)$ is divisible by $n$, and $f(n)-f(1)$ is divisible by $n-1$. Whence $-1$ is divisible by both $n$ and $n-1$ and you can complete the argument in various ways.
I mention the method because it can be used to answer questions about $n$ in other circumstances. For example, if you are given/have $f(7)=3$ and you want $f(n)=0$ you have $(n-7)|3$ so that $n=4, 6, 8, 10$ are the only possibilities.
A: We know
$$f(0) \equiv 1 \pmod{2}$$
This means that the constant term of $f$ is an odd integer.

We also know 
$$f(1) \equiv 1 \pmod{2}$$
This means that the sum of the coefficients of $f$ is odd.

Consider $f(x)$ for a general $x \in \mathbb{Z}$. 
Case 1: $x$ is even.
If $x$ is even, all of the powers of $x$ must also be even, and since the product of even numbers with any integer coefficients is also even, everything but the constant term of $f$ is even. Thus, $f(x)$ must be odd. 
Case 2: $x$ is odd.
If $x$ is odd, all of the powers of $x$ must also be odd. We have a bunch of odd numbers $x^i$ (including $x^0)$, and we are adding multiples of them together. We know that the sum of the coefficients is odd; this means we have, in total, an odd number of $x^i$s. The sum of an odd number of odd numbers must be odd, and so $f(x)$ is odd. 

We've proven that $f(x)$ is odd for integer $x$. It follows, a fortiori, that $f(x)$ is nonzero.
