Bashing unavailable - part 3

Algebra Level 4

Given that

$\color{#20A900}{a+b+c =1}$

$\color{#3D99F6}{a^2+b^2+c^2=2}$

$\color{#D61F06}{a^3+b^3+c^3=3}$

Then $a^4+b^4+c^4= \frac{a_1}{a_2}$ .... where $a_1$ and $a_2$ are positive coprime integers.

And $a^5+b^5+c^5=a_3$ ...... where $a_3$ is an integer.

And $a^6+b^6+c^6= \frac{a_4}{a_5}$ ....where $a_4$ and $a_5$ are positive coprime integers.

Find $a_1+a_2+a_3+a_4+a_5$ .

The answer is 152.

2 solutions

John Ashley Capellan
Jun 7, 2014

Using Newton's Identity where:

$a(n)S(1) + a(n-1) = 0$

$a(n)S(2) + a(n-1)S(1) + 2a(n-2) = 0$

$a(n)S(3) + a(n-1)S(2) + a(n-2)S(1) + 3a(n-3) = 0$

$a(n)S(4) + a(n-1)S(3) + a(n-2)S(2) + a(n-3)S(1) + 4a(n-4) = 0$

$a(n)S(5) + a(n-1)S(4) + a(n-2)S(3) + a(n-3)S(4) + a(n-4)S(5) + 5a(n-5) = 0$

$a(n)S(6) + a(n-1)S(5) + a(n-2)S(4) + a(n-3)S(3) + a(n-4)S(2) + 5a(n-5) = 0$

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 $a(n)S(6) + a(n-1)S(5) + a(n-2)S(4) + a(n-3)S(3) = 0$

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 $S(6) = 103/12$ and from earlier, $S(5) = 6$ and $S(4) = 25/6$ . Hence, the answer.

Rishabh Tiwari
Apr 25, 2016

It would be much easier if u use mr. Sanjeet raria's note on newtons sums...by the way nice solution.

Bashing unavailable - part 3

The answer is 152.

2 solutions

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