Solutions of 2016

Let $g(x)$ be a polynomial with integer coefficients such that $g(x)=f(x)-2015$ .

Then, $g(a_1)=g(a_2)=g(a_3)=g(a_4)=g(a_5)=0$ , since $f(a_i)=2015$ for $i=1,2,3,4,5$ . This implies that the integers $a_i$ are 5 distinct roots of the equation $g(x)=0$ . We can rewrite $g(x)$ as

$g(x)=k(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)h(x)$

where $k$ is a integer and $h(x)$ is another polynomial with integer coefficients.

Let us assume that $f(n)=2016$ for some integer $n$ . Then, we will have

$g(n)=k(n-a_1)(n-a_2)(n-a_3)(n-a_4)(n-a_5)h(n)=f(n)-2015=1$

Now, all the $a_i$ s being 5 distinct integers, $(n-a_i)$ s are also 5 distinct integers. Also, $h(n)$ is an integer because $h(x)$ is a polynomial with integer coefficients. Hence, the above result equates 1 to the product of 7 integers, of which at least 5 are distinct. This is not possible since 1 is a co-prime number, which is divisible only by itself.

Therefore, there are no integral solutions of the equation $f(x)=2016$ .