Divisible by 1972

$1972=2^2*17*29$ So if a number is divisible by all of the $4, 17, 29$ numbers, then it is divisible by $1972$ .

We wil use that $a-b\mid a^n-b^n$ for any $a, b, n$ positive integers.

Divisible by $17$ :

$S=(345^n-5^n)+(179^n-26^n)+493^n$

Since $\begin{aligned} 345-5=340 & = 20*17 & \mid 345^n-5^n \\ 179-26=153 & = 9*17 & \mid179^n-26^n \\ 493 & = 17*29 & \mid 493^n\end{aligned}$ $S$ is divisible by $17$ .

Divisible by $29$ :

$S=(345^n-26^n)+(179^n-5^n)+493^n$

Since

$\begin{aligned} 345-26=319 & = 11*29 & \mid 345^n-26^n \\ 179-5=174 & = 6*29 & \mid179^n-5^n \\ 493 & = 17*29 & \mid 493^n \end{aligned}$ $S$ is divisbly by $29$ .

Divisible by $4$ :

It is clear that $S$ always divisible by $2$ , because the sum of three odd numbers - an even and an odd number = even number. The question is that what is the remainder when it is divided by $4$ .

$S=(4*86+1)^n+(45*4-1)^n-(4+1)^n-(6*4+2)^n+(123*4+1)^n$

Note that $(4k+l)^n\equiv l^n \ \mod 4$ This can be proved using the Binominal theorem. From that $\begin{aligned} (4*86+1)^n & \equiv 1 \ \mod4 \\ (45*4-1)^n & \equiv \begin{cases} \text{If n is odd}\Longrightarrow -1 \\ \text{If n is even} \Longrightarrow 1 \end{cases} \ \mod4 \\ (4+1)^n & \equiv 1 \ \mod 4 \\ (6*4+2)^n & \equiv \begin{cases} n=1 \Longrightarrow 2 \\ n>1 \Longrightarrow 0 \end{cases} \\ (123*4+1)^n & \equiv 1 \ \mod 4 \end{aligned}$

So the remainders: $\begin{aligned} n=1\space & S & \equiv 2 \ \mod 4 \\ \text{n is odd}\quad n>1\space & S & \equiv 0 \ \mod 4 \\ \text{n is even}\space & S & \equiv 2 \ \mod 4 \end{aligned}$

So $S$ will divisible by $1972$ , if (and only if) $n$ is an odd number, $>1$ .

Therefore the answer is $\boxed{1008}$ .

[Not a solution.]

By trying sufficient $n$ , we can conjecture that $1972$ only divides $S$ if $n$ is odd and $\geq 3$ , making the answer $\dfrac{2017 - 3}{2} + 1 = \boxed{1008}$ assuming that our conjecture is in fact true for all such $n$ .