More fun in 2016, Part 9

Number Theory Level 5

Let s ( n ) s(n) be the sum of the n n th powers of all the primitive 2016 2016 th roots of unity, ω \omega . Find the minimal value of s ( n ) s(n) for all positive integers n n .


The answer is -576.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Andreas Wendler
Jan 2, 2016

Since primitive 2016th roots {k} occur pairwise conjugate complex sought sum S can be written:

S = 2 ∑ { k } 1007 c o s ( 2 π k n 2016 ) S=2\sum_{\{k\}}^{1007}cos(\frac{2 \pi kn}{2016})

For n=1008 the argument of cosinus is π k \pi k with odd numbers k. I.e. each summand (for half primitive roots) delivers minimum possible value -1. This however means the sum is

− ϕ ( 2016 ) = − 2 5 ( 1 − 1 2 ) 3 2 ( 1 − 1 3 ) ( 7 − 1 ) = − 576 - \phi(2016)=-2^{5}(1-\frac{1}{2})3^{2}(1-\frac{1}{3})(7-1)= -576

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Ja, genau! (+1) I did not even think of considering conjugates, but I just observed that if ω \omega is a primitive 2016th root of unity, then ω 1008 \omega^{1008} is a primitive second root of unity, so that ω 1008 = − 1 \omega^{1008}=-1 . Adding them up, we find the minimal value s ( 1008 ) = ϕ ( 2016 ) × ( − 1 ) = − 576 s(1008)=\phi(2016)\times(-1)=-576 .

Shall we try the same problem with an odd number?

Ich wünsche Ihnen auch ein wunderbares Neues Jahr, gute Gesundheit, viel Spass mit der Mathematik und Fa­mi­li­en­glück.

Otto Bretscher - 5 years, 5 months ago

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