Let \(\displaystyle{A}={\left({2}{\sin{{\frac{{{7}\pi}}{{{18}}}}}}+{1}\right)}^{{{2556}}}\) How can one find the remainder

siliciooy0j

siliciooy0j

Answered question

2022-03-24

Let A=(2sin7π18+1)2556
How can one find the remainder from the division of [A] by 17?

Answer & Explanation

strahujufl75

strahujufl75

Beginner2022-03-25Added 8 answers

2sin7π18+1=2cosπ9+1
B(n)=(1+2cosπ9)n+(1+2cos5π9)n+(1+2cos7π9)nZ
Try to find the remainders modulo 17, and do they repeat?
Can you find a recursion for B(n)? Hint: the three bases in B(n) are the roots of x33x2+1=0

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