Zeta gcd

gcd ( n , 2016 ) = 1 1 n 4 = a b π c \large \sum_{\gcd(n,2016)=1} \frac{1}{n^{4}} = \dfrac{a}{b}\pi^{c}

The equation above holds true for positive integers a , b a,b and c c with a , b a,b coprime. Find a + b + c a+b+c .

Notation : gcd \gcd denotes the greatest common divisor function.


Similar Problem .


The answer is 196485.

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1 solution

Otto Bretscher
Apr 19, 2016

Nice problem, Comrade! The proof of the Euler product formula offers a one-step solution:

S = ( 1 1 2 4 ) ( 1 1 3 4 ) ( 1 1 7 4 ) ζ ( 4 ) = 2000 π 4 194481 S=\left(1-\frac{1}{2^4}\right)\left(1-\frac{1}{3^4}\right)\left(1-\frac{1}{7^4}\right)\zeta(4)=\frac{2000\pi^4}{194481} . The answer is 196485 \boxed{196485}

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