Bashing unavailable - part 4
⎩ ⎪ ⎨ ⎪ ⎧ a + b + c = 1 a 2 + b 2 + c 2 = 2 a 3 + b 3 + c 3 = 3
Let a , b and c be complex numbers satisfying the system of equations above.
Given that the following equations are true as well for positive integers a 1 , a 2 , a 3 , a 4 , a 5 , a 6 and a 7 , with g cd ( a 1 , a 2 ) = g cd ( a 3 , a 4 ) = 1 = g cd ( a 6 , a 7 ) = 1 .
Find a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 .
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a 4 + b 4 + c 4 = a 2 a 1 a 5 + b 5 + c 5 = a 3 a 6 + b 6 + c 6 = a 5 a 4 a 7 + b 7 + c 7 = a 7 a 6
The answer is 391.
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Using Newton's Identity...
But since the polynomial is of third-degree polynomial, there are 4 terms and since there are 4 terms, there are 4 numerical coefficients, so for a(n-k) = 0 for k greater than or equal to 4.
This reduces to a(n)S(7) + a(n-1)S(6) + a(n-2)S(5) + a(n-3)S(4) = 0 where from earlier, we assumed that a(n) = 1, so a(n-1) = -1, a(n-2) = -1/2, and a(n-3) = -1/6.
Where we assume that a(n) = leading coefficient of polynomial = 1 of a(3)x^3 + a(2)x^2 + a(1)x + a(0) = P(x).
After algebraic manipulations, we get that S(7) = 221/18 and from earlier, S(6) = 103/12, S(5) = 6, and S(4) = 25/6. Hence, the answer.