2019 is Product of Two Primes

$2019$ have the following divisors: $\{1, 3, 673, 2019 \}$ . So, every integer $n$ except for multiples of $3$ and $673$ have $\gcd(n,2019) = 1$ .

Thus, we have to take the sum over all integer $n$ and then subtract $n$ multiple of $3$ or $673$ , i.e., $n = 3k$ or $n=673k$ . But by doing so we subtract two times the multiples of both $3$ and $673$ , i.e., the multiples of $3 \cdot 673 = 2019$ , or $n = 2019k$ . So we add them back. Then:

$\large \displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k^2} - \sum_{k=1}^{\infty} \frac{1}{(3k)^2} - \sum_{k=1}^{\infty} \frac{1}{(673k)^2} + \sum_{k=1}^{\infty} \frac{1}{(2019k)^2}$

$\large \displaystyle S = \left ( 1 - \frac{1}{3^2} - \frac{1}{673^2} + \frac{1}{2019^2} \right ) \sum_{k=1}^{\infty} \frac{1}{k^2}$

$\large \displaystyle S = \frac{1207808}{1358787} \cdot \frac{\pi^2}{6}$

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{603904 \pi^2}{4076361} }$

So:

$\large \displaystyle \color{#3D99F6} a = 603904, b = 2, c = 4076361, \boxed{\large \displaystyle a+b+c = 4680267}$

$\large S = \sum_{n \nmid 3,673 } \frac{1}{n^2} = \prod_{p \text{ is prime} \neq 3,673} \sum_{n=0}^{\infty} \frac{1}{p^{2n}} = \prod_{p \text{ is prime}} \left( 1 - p^{-2}\right)^{-1} \left( 1 - 3^{-2}\right)\left(1 - 673^{-2} \right) \\ \displaystyle \large= \zeta(2) \left( 1 - 3^{-2}\right)\left(1 - 673^{-2} \right) = \boxed{\pi^2\cdot\frac{603904}{4076361}}$