Polynomial with no integer roots This is an excercise given to a kid I am tutoring, as part of a set of problems regarding polynomials. He is currently at the last class before graduation year. 
Let $p$ be a polynomial in $ℤ[x]$ such that both $p(2015)$ and $p(2016)$ are odd numbers. Show that the polynomial has no integer roots.  
My approach is as follows:We will answer the question in a more general form. Let $p$ be a polynomial in $ℤ[x]$ such that there exists an integer $a$, so that both $p(a)$ and $p(a+1)$ are odd .
If $p$ has no roots in $ℝ$, then obviously it has no roots in $ℤ$.
Let $r_i$ be its roots and at least one of the $r_i$ belongs to $ℤ$. Let $r_j \inℤ$. Then we can write $p$ in the form: $p(x)=(x-r_1)^{n_1}(x-r_2)^{n_2}..(x-r_m)^{n_m}$  Since both $p(a)$ and $p(a+1)$ are odd it must follow that $(a-r_j)$ and $(a+1-r_j)$ are odd since if either one is even then the product would also be even. (Showing that the $n_j$-th power of an even or odd is also even or odd is trivial) But that is a contradiction, for the sums of a given number with an odd and an even number respectively cannot be both odd or both even. It follows that $p$ cannot have an integer root.I think this approach works but I am not at all sure if this solution is what their teacher had in mind. I find this rather embarassing, despite still being a uni undergrad myself, but I can't come up with something better, or rather bettter suited for the situation. Any ideas will be welcome.
 A: I'm not convinced by your argument: why can't one of $(a - r_j)$ or $(a + 1 - r_j)$ be even and have the even-ness cancelled by a [power of] $2$ in the denominator of some other factor?
Consider instead checking modulo $2$: if $f(n) = 0$, then $f(n)\equiv 0\pmod{2}$. However, if you're already looking at the reduced $\overline{f}\in\mathbb{F}_2[x]$, then you only have two things to plug in ($0$ and $1$). However, you already know $\overline{f}(0)$ and $\overline{f}(1)$: and they're both $1$.
A: As Stahl already pointed out, your proof doesn't quite work. What you need to do is say: suppose $p$ has an integer root $n$, then we have a factorization $p(x)=(x-n)q(x)$ for some $q\in\mathbb{Z}[X]$. Then you can deduce that $x-n$ must be even for either 2015 or 2016 and the evenness of $p(2015)$ respectively $p(2016)$ would follow..
Alternately, you can do the modulo two proof in a disguised form: use the two expansions $p(x)=\sum_{i=0}^ka_ix^i=\sum_{i=0}^kb_i(x-1)^i$; since the sum of even numbers is even, $p(2016)$ being odd implies $a_0$ being odd and $p(2015)$ being odd implies $b_0$ being odd. Conversely, if $n$ is even, then $a_0$ being odd implies $p(n)$ is odd and if $n$ is odd, then $b_0$ being odd implies $p(n)$ is odd - either way $p(n)$ is odd so in particular $p(n)$ must be non-zero for any integer $n$.
