So Many Squares
( x + x 1 ) 2 + ( x 2 + x 2 1 ) 2 + ( x 3 + x 3 1 ) 2 + … + ( x 1 0 0 9 + x 1 0 0 9 1 ) 2
What is the value of the expression above if x + x 1 = − 1 ?
The answer is 2017.
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We can start by some calculations: x + 1/x = -1 => (x + 1/x)^2 = 1 x^2 + 1/x^2 = (x + 1/x)^2 - 2 = -1 => (x^2 + 1/x^2)^2 = 1 x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = (-1)^3 - 3(-1) = 2 => (x^3 + 1/x^3)^2 = 4 x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2 = 1 - 2 = -1 => (x^4 + 1/x^4)^2 = 1 x^5 + 1/x^5 = (x^4 + 1/x^4)(x + 1/x) - (x^3 + 1/x^3) = (-1)(-1) - 2 = -1 => (x^5 + 1/x^5)^2 = 1 x^6 + 1/x^6 = (x^3 + 1/x^3)^2 - 2 = 2^2 - 2 = 2 => (x^6 + 1/x^6)^2 = 4
So we can predict that x^n + 1/x^n: (n is a positive integer) 1/ = 2 if n is divisible by 3 2/ = -1 if n is not divisible by 3 (*)
Proved by induction: 1/ x^3 + 1/x^3 = x^6 + 1/x^6 = 2 Provided that there exists n = 3k that x^3k + 1/x^3k = 2, we will prove that x^3(k+1) + 1/x^3(k+1) = 2 x^3(k+1) + 1/x^3(k+1) = (x^3k + 1/x^3k)(x^3 + 1/x^3) - [x^3(k-1) + 1/x^3(k-1)] = 2*2 - 2 = 2 (true)
2/ x + 1/x = x^2 + 1/x^2 = x^4 + 1/x^4 = x^5 + 1/x^5 = -1 Provided that there exists n = 3k + 1 that x^(3k+1) + 1/x^(3k+1) = -1, we will prove that x^(3(k+1)+1) + 1/x^(3(k+1)+1) = -1 x^(3k+4) + 1/x^(3k+4) = x^(3k+1) + 1/x^(3k+1) - [x^(3(k-1)+1) + 1/x^(3(k-1)+1)] = (-1)*2 - (-1) = -1 (true) The same with n = 3k + 2
From (*) we have (x^n + 1/x^n)^2: 1/ = 4 if n is divisible by 3 2/ = 1 if n is not divisible by 3
3, 6, 9, …, 1008 are divisible by 3 => 336 numbers are divisible by 3 and 1009 - 336 = 673 aren't SUM = 336 4 + 673 1 = 2017