An algebra problem by Benny Sampson Uche
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x + y + z = 1 x 2 + y 2 + z 2 = 2 x 3 + y 3 + z 3 = 3 x 4 + y 4 + z 4 = n m
The numbers x , y , and z satisfy the system of equations above, where m and n are coprime positive integers. Find m + n .
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3 solutions
x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) + 3 x y z . So 3 − 3 x y z = 2 − ( − 1 / 2 ) , x y z = 1 / 6 . Thus x 4 + y 4 + z 4 = 7 / 2 + 4 x y z = 7 / 2 + 4 / 6 = 2 5 / 6 . The answer is 2 5 + 6 = 3 1
You should change then language of the q as sum(cyclic)x^2(y^2)=-1/12 not possible for reals
Let P n = x n + y n + z n , where n ∈ N , and S 1 = x + y + z = 1 , S 2 = x y + y z + z x and S 3 = x y z . Using Newton's sum method , we have:
\(\begin{array} {} P_1 = S_1 & \implies P_1 = 1 \\ P_2 = S_1P_1 - 2S_2 = 1-2S_2 = 2 & \implies S_2 = -\frac 12 \\ P_3 = S_1P_2 - S_2P_1 + 3S_3 = 2 + \frac 12 + 2S_3 = 3 & \implies S_3 = \frac 16 \\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 = 3 + 1 + \frac 16 & \implies P_4 = x^4+y^4+z^4 = \frac {25}6 \end{array} \)
⟹ m + n = 2 5 + 6 = 3 1