Hidden Polynomial! (No easy logic)

Algebra Level 2

{ a + b + c = 3 a 2 + b 2 + c 2 = 5 a 3 + b 3 + c 3 = 7 \large \begin{cases}{a+b+c=3} \\ {a^2 +b^2 +c^2 =5} \\ {a^3 + b^3 +c^3 = 7} \end{cases}

If a , b , c a,b,c are complex numbers satisfying the system of equations above, find the value of a 4 + b 4 + c 4 a^4 + b^4 + c^4 .

Note: I found the problem in my Book. See: "Ayo Raih Medali Emas Olimpiade Matematika" (English: Let's Get A Gold Medal in Mathematics Olympiad) by: Dr. Khow Yao Tung, M. Sc. Ed, MEd.


The answer is 9.

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Figel Ilham by Figel Ilham , 22, Indonesia

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5 solutions

Figel Ilham
May 26, 2015

Let's we make the monic polynomial with degree 3, rewrite x 3 + A x 2 + B x + C x^3+ Ax^2 +Bx+C and let a , b , c a,b,c roots of the polynomial, then according to Newton's Sums with P k = a k + b k + c k P_k = a^k + b^k + c^k , then we have P 1 + A = 0 P_1 + A = 0 3 + A = 0 ⇒ A = − 3 3+A=0 \Rightarrow A = -3 P 2 + A P 1 + 2 B = 0 P_2 + AP_1+2B = 0 5 + ( − 3 ) 3 + 2 B = 0 ⇒ B = 2 5+(-3)3+2B=0 \Rightarrow B = 2 P 3 + A P 2 + B P 1 + 3 C = 0 P_3+AP_2+BP_1+3C = 0 7 + ( − 3 ) 5 + 2 ( 3 ) + 3 C = 0 ⇒ C = 2 3 7+(-3)5+2(3)+3C=0 \Rightarrow C = \frac{2}{3} P 4 + A P 3 + B P 2 + C P 1 = 0 P_4+AP_3+BP_2+CP_1 = 0 P 4 + ( − 3 ) 7 + 2 ( 5 ) + 2 3 ( 3 ) = 0 ⇒ P 4 = 9 P_4+(-3)7 +2(5)+\frac{2}{3}(3) = 0 \Rightarrow P_4 = 9

Note: using algebraic manipulation is extremely tedious!

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Moderator note:

Good. Bonus question: Is it true that a 5 + b 5 + c 5 = 11 a^5 + b^5 + c^5 = 11 ?

For Challenge Master note, it is absolutely INCORRECT. Using Newton's Sum, we have P 5 − 3 P 4 + 2 P 3 + 2 3 P 2 = 0 P_5 -3P_4 +2P_3+\frac{2}{3}P_2 = 0 The P 2 = 5 P_2 = 5 , since 2 3 × 5 \frac{2}{3}\times 5 is a fraction, absolutely the answer must be in fraction. Thus, observation is extremely a complete mistake! The result of P 5 = 29 3 P_5 = \frac{29}{3}

Thanks to Master @Pi Han Goh

Figel Ilham - 6 years ago

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It's 29 3 \frac{29}3 , not 49 3 \frac{49}{3} . EDIT: Haha, don't call me master~!

Pi Han Goh - 6 years ago

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Sorry, miscounted. Edited

Figel Ilham - 6 years ago

why is this question given in complex numbers search it should be there in vieta's .There is no trace of complex numbers in the solution.

Raven Herd - 3 years, 9 months ago

A good solution by Figel Ilham.

Mahavishnu Brandao - 6 years ago
Sujit Debnath
May 27, 2015

Given that, a + b + c a + b + c = 3, a 2 + b 2 + c 2 a^2 + b^2 + c^2 = 5, a 3 + b 3 + c 3 a^3 + b^3 + c^3 = 7

We know that,
formula-1: a 2 + b 2 + c 2 a^2 + b^2 + c^2 = ( a + b + c ) 2 − (a + b + c)^2 - 2 ( a b + b c + c a ) 2(ab + bc + ca)
formula-2: a 3 + b 3 + c 3 − 3 a b c a^3 + b^3 + c^3 - 3abc = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)

Using formula-1, ab + bc + ca = 3 2 − 5 2 \frac {3^2 - 5}{2} = 2
Using formula-2, abc = − 2 3 - \frac {2}{3}

From formula-1, we can write that,
( a b ) 2 + ( b c ) 2 + ( c a ) 2 (ab)^2 + (bc)^2 + (ca)^2
= ( a b + b c + c a ) 2 − 2 ( a b × b c + b c × c a + c a × a b ) (ab + bc + ca)^2 - 2(ab \times bc + bc \times ca + ca \times ab)
= ( a b + b c + c a ) 2 − 2 a b c ( a + b + c ) (ab + bc + ca)^2 - 2abc(a + b + c)
= 2 2 − 2 × ( − 2 3 ) × 3 2^2 - 2 \times (- \frac {2}{3}) \times 3 = 8

So, a 4 + b 4 + c 4 a^4 + b^4 + c^4
= ( a 2 ) 2 + ( b 2 ) 2 + ( c 2 ) 2 (a^2)^2 + (b^2)^2 + (c^2)^2
= ( a 2 + b 2 + c 2 ) 2 − 2 [ ( a b ) 2 + ( b c ) 2 + ( c a ) 2 ] (a^2 + b^2 + c^2)^2 - 2[(ab)^2 + (bc)^2 + (ca)^2]
= 5 2 − 2 × 8 5^2 - 2 \times 8 = 9

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Moderator note:

Yes, Newton's Identities is more than enough to solve this problem. Bonus question: Can you prove that a 5 + b 5 + c 5 = 29 3 a^5 + b^5 + c^5 = \frac{29}3 with just Newton's Identities alone?

Good ones!

Figel Ilham - 6 years ago

Good job :-)

Mehedi Hasan - 6 years ago

Prove of bonus question,
( a 2 + b 2 + c 2 ) ( a 3 + b 3 + c 3 ) (a^2 + b^2 + c^2)(a^3 + b^3 + c^3) = 5 × 7 5 \times 7
⇒ a 5 + a 2 b 3 + c 3 a 2 + a 3 b 2 + b 5 + b 2 c 3 + c 2 a 3 + b 3 c 2 + c 5 \Rightarrow a^5 + a^2b^3 + c^3a^2 + a^3b^2 + b^5 + b^2c^3 + c^2a^3 + b^3c^2 + c^5 = 35
⇒ a 5 + b 5 + c 5 + a 3 b 2 + a 2 b 3 + b 3 c 2 + b 2 c 3 + c 3 a 2 + c 2 a 3 \Rightarrow a^5 + b^5 + c^5 + a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3 = 35
⇒ a 5 + b 5 + c 5 + a 2 b 2 ( a + b ) + b 2 c 2 ( b + c ) + c 2 a 2 ( c + a ) \Rightarrow a^5 + b^5 + c^5 + a^2b^2(a + b) + b^2c^2(b + c) + c^2a^2(c + a) = 35
⇒ a 5 + b 5 + c 5 + a 2 b 2 ( 3 − c ) + b 2 c 2 ( 3 − a ) + c 2 a 2 ( 3 − b ) \Rightarrow a^5 + b^5 + c^5 + a^2b^2(3 - c) + b^2c^2(3 - a) + c^2a^2(3 - b) = 35
⇒ a 5 + b 5 + c 5 + 3 a 2 b 2 − a 2 b 2 c + 3 b 2 c 2 − a b 2 c 2 + 3 c 2 a 2 − a 2 b c 2 \Rightarrow a^5 + b^5 + c^5 + 3a^2b^2 - a^2b^2c + 3b^2c^2 - ab^2c^2 + 3c^2a^2 - a^2bc^2 = 35
⇒ a 5 + b 5 + c 5 + 3 [ ( a b ) 2 + ( b c ) 2 + ( c a ) 2 ] − a b c ( a b + b c + c a ) \Rightarrow a^5 + b^5 + c^5 + 3[(ab)^2 + (bc)^2 + (ca)^2] - abc(ab + bc + ca) = 35
⇒ a 5 + b 5 + c 5 + 3 × 8 − ( − 2 3 ) × 2 \Rightarrow a^5 + b^5 + c^5 + 3 \times 8 - (- \frac {2}{3}) \times 2 = 35
So, a 5 + b 5 + c 5 a^5 + b^5 + c^5 = 29 3 \frac {29}{3}

Sujit Debnath - 6 years ago
Rajen Kapur
Jun 2, 2015

I offer appreciation for the Note in the question, as it reveals the beauty of Indonesian language.

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What makes so special :) ?

Figel Ilham - 6 years ago

each and every step increase 2 (value) ) ..so finally we get answer is 7 ![](

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Moderator note:

Like Ahmed Karrany's solution, your logic is flawed. Is Figel Illham's solution.

See my solution. It's flawed

Figel Ilham - 6 years ago
Ahmed Karrany
May 26, 2015

by observation

the sum of a, b and c=3 ...........(1)

the sum of square of each a, b and c=5...........(2)

the sum of cube of each a, b and c=7 ...........(3)

the diffrence between Eqs 2 & 1 is 2

the diffrence between Eqs 3 & 2 is 2 also.

Hence

the sum of fourth order of each a, b and c=9 ...........(4)

the diffrence between Eqs 4 & 3 is 2 (constant).

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Moderator note:

You're still wrong. You can't just say the pattern works because you noticed it applies for small values of n n .

If I replaced the numbers 3 , 5 , 7 3,5,7 by 1 , 2 , 3 1,2,3 respectively, do you still think that the answer will be 4 4 ? No!

Your logic is flawed. See Figel Ilham's solution.

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