If x 2 + x + 1 = 0 , then r = 1 ∑ 2 5 ( x r + x r 1 ) 2 is equal to :
ω is non-real cube root of unity.
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Can you suggest some place from where I can learn Modular Arithmetic ? I had also tried reading from Brilliant but it was a bit too advanced for me .
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You can try from khanacedemy , as you can learn in videos, you may feel easy.
x 2 + x + 1 = 0 t h e n x = 2 1 ± i 3 = c o s 6 0 + i s i n 6 0 = e i 6 0 ∴ ( x + x 1 ) = 2 c o s 6 0 f r o m D e − m o i v r e ′ s t h e o r e m ( x n + x n 1 ) = 2 c o s ( 6 0 n ) ∴ ( x n + x n 1 ) 2 = 1 i f n m o d 3 ≡ 1 o r 2 , ( x n + x n 1 ) 2 = 4 i f n m o d 3 ≡ 0 . ∴ ∑ n = 1 2 5 ( x n + x n 1 ) 2 = 4 × 4 + 4 × 8 + 1 = 4 9 .
Using the Euler identity:
e i θ = cos θ + i sin θ and e − i θ = cos θ − i sin θ
we find:
e i θ + e − i θ = 2 cos θ
for x 2 + x + 1 = 0 → x = 2 1 ± i 3 = e ± i 3 π
Now we simply plug and chug:
r = 1 ∑ 2 5 ( e i 3 π r + e i 3 π r 1 ) 2 = r = 1 ∑ 2 5 ( e i 3 π r + e − i 3 π r ) 2 = r = 1 ∑ 2 5 ( 2 cos 3 π r ) 2
or:
r = 1 ∑ 2 5 ( e i 3 π r + e − i 3 π r ) 2 = r = 1 ∑ 2 5 ( e i 3 2 π r + e − i 3 2 π r + 2 ) = r = 1 ∑ 2 5 ( 2 cos 3 2 π r + 2 )
r = 1 ∑ 2 5 ( 2 cos 3 2 π r + 2 ) = 2 ( r = 1 ∑ 2 5 cos 3 2 π r + 2 5 )
when r is not a multiple of 3: cos 3 2 π r = − 2 1 (which is a set of 17 possible numbers)
when r is a multiple of 3: cos 3 2 π r = 1 (which is a set of 8 possible numbers)
∴ 2 ( r = 1 ∑ 2 5 cos 3 2 π r + 2 5 ) = 2 ( 1 7 ( 2 − 1 ) + 8 ( 1 ) + 2 5 ) = 4 9
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( ω r + ω r 1 ) 2 = 1 if r ≡ 1 , 2 m o d 3 ( ω r + ω r 1 ) 2 = 4 if r ≡ 0 m o d 3
Thus, the answer will be 8 × 4 + 1 × 1 7 = 4 9 , as there are 8 multiples of 3 in [ 1 , 2 5 ]