Minimizing degrees!

Let $f(x)= (x+a)(x+a+1)(x+a+2)(x+a+3)$ for any integer $a$ . Then, in any given four consecutive integers, exactly two of them are divisible by 2 and exactly one of them is divisible by 4, so their product is always divisible by 8. The degree of $f(x)$ is then $\boxed{4}$

We then try to do better than this. Suppose a cubic polynomial $p(x)$ of the form $x^3+ax^2+bx+c$ satisfies. Obviously, $p(0)$ must satisfy, so $(0)^3+a(0)^2+b(0)+c \equiv c \equiv 0 \pmod{8}$ . So, $c \equiv 0 \pmod{8}$ .

We continue working mod 8. We see that $p(1) = 1+a+b+c \equiv 1+a+b \pmod {8} \equiv 0 \pmod {8}$ , so $a+b \equiv -1 \pmod{8}$ .

But then, $p(-1) = -1+a-b+c \equiv -1+a-b \pmod {8} \equiv 0 \pmod{8}$ , so $b-a \equiv -1 \pmod {8}$ .

From here we can deduce that $a \equiv 0 \pmod{8}$ , $b \equiv -1 \pmod{8}$ . So, taking $p(x) \pmod {8}$ ,

$p(x) = x^3+ax^2+bx+c \equiv (x^3-x) \pmod 8$ . We then see that $p(2)$ does not satisfy $p(2) \equiv 0 \pmod{8}$ , thus a contradiction. We can also follow the steps above to disprove the possibility of $p(x)$ being quadratic. In fact, paragraphs 3 to 5 have already done so, only that all the terms are multiplied with $x$ . Just divide by $x$ and the proof for the quadratic case is done.

Finally, if $p(x)$ is linear, then $p(x)=x+c$ , which must be divisible by 8 for all $x$ , which is absurd. Therefore, only quartic equations satisfy.

The function on integers gives an output divisible by 8. then starting from one integer n, n(n+1) would already be divisibly by 2 no matter what n will be. n(n+1)(n+2) would be divisible by 8 only when n is even. So to ensure the function for all integers, we have n(n+1)(n+2)(n+3), making the function a fourth-degree polynomial.