1,2,3,will this continue?

Algebra Level 3

{ a + b + c = 1 a 2 + b 2 + c 2 = 2 a 3 + b 3 + c 3 = 3 \large \begin{cases} \begin{aligned} a\ + \ b\ +\ c \ & =1 \\ a^2+b^2+c^2 & =2 \\ a^3+b^3+c^3 & =3 \end{aligned} \end{cases}

If a a , b b , and c c satisfy the system of equations above, find the value of a 4 + b 4 + c 4 4 a^4+b^4+c^4-4 .

1 2 \frac{1}{2} 1 3 \frac{1}{3} 0 0 1 6 \frac{1}{6}

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2 solutions

Chew-Seong Cheong
Dec 16, 2018

Relevant wiki: Newton's Identities

Let P n = a n + b n + c n P_n = a^n+b^n+c^n , where n n is a positive integer, and S 1 = a + b + c = 1 S_1 = a+b+c=1 , S 2 = a b + b c + c a S_2 = ab+bc+ca , and S 3 = a b c S_3 = abc . By Newton's sums or Newton's identities:

P 1 = S 1 = 1 P 2 = S 1 P 1 2 S 2 = 1 ( 1 ) 2 S 2 = 2 S 2 = 1 2 P 3 = S 1 P 2 S 2 P 1 + 3 S 3 = 1 ( 2 ) + 1 2 ( 1 ) + 3 S 3 = 3 S 3 = 1 6 P 4 = S 1 P 3 S 2 P 2 + S 3 P 1 = 1 ( 3 ) + 1 2 ( 2 ) + 1 6 ( 1 ) = 4 1 6 \begin{aligned} P_1 & = S_1 = 1 \\ P_2 & = S_1P_1 - 2S_2 = 1(1) - 2S_2 = 2 & \small \color{#3D99F6} \implies S_2 = - \frac 12 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 1(2) + \frac 12 (1) + 3S_3 = 3 & \small \color{#3D99F6} \implies S_3 = \frac 16 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 1(3) + \frac 12 (2) +\frac 16(1) = 4\frac 16 \end{aligned}

Therefore, P 4 4 = 1 6 P_4 - 4 = \boxed{\dfrac 16} .

If ( P i = x i + y i + z i ) (P_i=x^i+y^i+z^i)

According to Newton's identities,

P 4 = 4 P 3 P 1 3 P 2 P 1 2 + P 1 4 6 + P 2 2 2 P_4=\frac{4P_3P_1}{3}-P_2{P_1}^2+\frac{{P_1}^4}{6}+\frac{{P_2}^2}{2}

so we get P 4 = 4 3 1 3 2 1 2 + 1 4 6 + 2 2 2 = 4 + 1 6 . P_4=\frac{4*3*1}{3}-2*{1}^2+\frac{{1}^4}{6}+\frac{{2}^2}{2} = 4 + \frac{1}{6}.

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