Functionally, a Number theory problem

The solution given by priyanshu mishra is not the most elegant solution. since we have f(mn)+1 = (f(m)+1)(f(n)+1), Let me denote statement S that states: "f(x)+1 is divisible by x for all integers x." Now let us assume statement S true. It implies that f(m)+1 is divisible by m and f(n)+1 is divisible by n. so (f(m)+1)(f(n)+1) is divisible by mn. Since this is f(mn)+1 , we also have f(mn)+1 is divisible by mn. Hence statement S is true. Now, since f(2017) +1 is divisible by 2017, we have f(2017) is congruent to 2016 modulo 2017. Hence 2016 is the remainder

Observe that $f(0) = f(1) = 0$ . For $n \ge 2$ , let us define $g(n) = f(n) + 1$ . Then $g(2) = 8$ and

$g(mn) = f(mn) + 1 = f(m) + f(n) + f(mn) + 1 = (f(m) + 1)(f(n) + 1) = g(m)g(n)$

Fix an integer $n > 2$ and consider a sequence ${ p }_{ k } / { q }_{ k }, k \ge 1$ , of rational numbers that are greater than $\log _{ 2 }{ n }$ and converge to $\log _{ 2 }{ n }$ ( here ${ p }_{ k }$ and ${ q }_{ k }$ are integers). Then from $\large\ n < { 2 }^{ { p }_{ k }/{ q }_{ k } }$ , so by monotonicity of $g$ ,

$\large\ g({ n }^{ { q }_{ k } })\le g({ 2 }^{ { p }_{ k } })$ .

The multiplicativity of $g$ implies that

$\large\ g(n)\le g{ \left( 2 \right) }^{ { p }_{ k }/{ q }_{ k } }={ 2 }^{ 3{ p }_{ k }/{ q }_{ k } }=\left( { \left( { 2 }^{ { p }_{ k }/{ q }_{ k } } \right) }^{ 3 } \right)$ .

Passing to the limit with $k\rightarrow \infty$ , we find that $g(n) \le {n}^3$ . A similar argument shows that $g(n) \ge {n}^3$ . Hence $g(n) = n^3$ , and so $f(n) = n^3 - 1$ for all $n$ is the unique solution to the functional equation.

$f(2017) \equiv2016 \pmod{2017}$ . So the answer is $\boxed{2016}$ .