Going back to the roots!

Algebra Level 4

Given that x 1 x_1 , x 2 x_2 , and x 3 x_3 are the roots of the polynomial f ( x ) = 4 x 3 − 24 x 2 + 44 x − 23 f(x)=4x^3-24x^2+44x-23 , you can express the number

( x 1 x 2 ) 4 + ( x 2 x 3 ) 4 + ( x 1 x 3 ) 4 (x_1 x_2)^4+(x_2 x_3 )^4+(x_1 x_3)^4

as a fraction p q \dfrac{p}{q} , where p p and q q are positive coprime integers. Find p + q p+q .


The answer is 7117.

Arturo Presa by Arturo Presa , 62, USA

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2 solutions

Arturo Presa
Dec 15, 2020

Using Vieta's formulas we get that x 1 + x 2 + x 3 = 6 , x 1 x 2 + x 2 x 3 + x 1 x 3 = 11 x_1+x_2+x_3= 6, x_1 x_2+x_2 x_3 + x_1 x_3= 11 and x 1 x 2 x 3 = 23 / 4. x_1 x_2 x_3 = 23/4. We can construct a new polynomial g ( x ) = x 3 + a x 2 + b x + c , g(x)= x^3+ax^2+bx+c, whose roots are x 1 x 2 , x 2 x 3 , x_1 x_2, x_2 x_3, and x 1 x 3 . x_1 x_3. Using Vieta's formulas again and the previous equations we get a = − ( x 1 x 2 + x 2 x 3 + x 1 x 3 ) = − 11 a=-(x_1 x_2+x_2 x_3 + x_1 x_3)=-11\quad\quad\quad\quad \quad\quad\quad\quad\quad b = x 1 2 x 2 x 3 + x 1 x 2 2 x 3 + x 1 x 2 x 3 2 = x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) = 23 4 × 6 = 69 2 \quad\quad\quad\quad \quad\quad b= x_1^2 x_2 x_3+x_1 x_2^2 x_3 + x_1 x_2 x_3^2= x_1 x_2 x_3 (x_1+x_2+x_3)=\frac{23}{4}\times6=\frac{69}{2} c = − ( x 1 x 2 ) ( x 2 x 3 ) ( x 1 x 3 ) = − ( x 1 x 2 x 3 ) 2 = − ( 23 4 ) 2 = − 529 16 . \quad c=-(x_1 x_2)(x_2 x_3) ( x_1 x_3)=-(x_1 x_2 x_3)^2=-(\frac{23}{4})^2=-\frac{529}{16}. So, the numbers x 1 x 2 , x 2 x 3 x_1 x_2, x_2 x_3 and x 1 x 3 x_1 x_3 are the roots of the polynomial g ( x ) = x 3 − 11 x 2 + 69 2 x − 529 16 . g(x)=x^3-11x^2+\frac{69}{2} x-\frac{529}{16}.

Now, let us define α n = ( x 1 x 2 ) n + ( x 2 x 3 ) n + ( x 1 x 3 ) n . \alpha_n= (x_1 x_2)^n+(x_2 x_3)^n +( x_1 x_3)^n. It is obvious that α 0 = 3 \alpha_0=3 (none of the x i ′ s x_i's is equal to zero) and α 1 = x 1 x 2 + x 2 x 3 + x 1 x 3 = 11. \alpha_1=x_1 x_2+x_2 x_3 + x_1 x_3=11. We can also find α 2 . \alpha_2. Indeed, α 2 = ( x 1 x 2 ) 2 + ( x 2 x 3 ) 2 + ( x 1 x 3 ) 2 = ( x 1 x 2 + x 2 x 3 + x 1 x 3 ) 2 − 2 x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) = 1 1 2 − 2 × 23 4 × 6 = 52. \alpha_2 = (x_1 x_2)^2+(x_2 x_3)^2 +( x_1 x_3)^2=(x_1 x_2+x_2 x_3 + x_1 x_3)^2-2 x_1 x_2 x_3 (x_1+x_2+x_3)=11^2- 2\times\frac{23}{4}\times 6=52. Now it is easy to see that α n \alpha_n is a linear recurrent sequence such that α n = 11 α n − 1 − 69 2 α n − 2 + 529 16 α n − 3 , \alpha_n= 11\alpha_{n-1}-\frac{69}{2} \alpha_{n-2}+ \frac{529}{16}\alpha_{n-3}, and α 0 = 3 , α 1 = 11 \alpha_0=3, \alpha_1= 11 and α 2 = 52. \alpha_2=52. Using the recurrence relation, we get that α 3 = 4667 16 \alpha_3=\frac{4667}{16} and α 4 = 7113 4 . \alpha_4=\frac{7113}{4}. Therefore, the answer is 7113 + 4 = 7117 . \boxed{7113+4=7117}.

1 Helpful 1 Interesting 4 Brilliant 0 Confused

An alternative to the recurrence would be r 1 4 + r 2 4 + r 3 4 = ( r 1 2 + r 2 2 + r 3 2 ) 2 − 2 ( ( r 1 r 2 ) 2 + ( r 2 r 3 ) 2 + ( r 3 r 1 ) 2 ) = ( ( r 1 + r 2 + r 3 ) 2 − 2 ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) ) 2 − 2 ( ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) 2 − 2 ( r 1 + r 2 + r 3 ) ( r 1 r 2 r 3 ) ) r_1^4+r_2^4+r_3^4=(r_1^2+r_2^2+r_3^2)^2-2((r_1r_2)^2+(r_2r_3)^2+(r_3r_1)^2)=((r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1))^2-2((r_1r_2+r_2r_3+r_3r_1)^2-2(r_1+r_2+r_3)(r_1r_2r_3)) where r 1 = x 1 x 2 , r 2 = x 2 x 3 r_1=x_1x_2, r_2=x_2x_3 and r 3 = x 3 x 1 r_3=x_3x_1

Sathvik Acharya - 5 months, 4 weeks ago

How was 69/2 gotten

Odijie Dayvhid - 5 months, 4 weeks ago

Log in to reply

It is the value of a . a.

Arturo Presa - 5 months, 3 weeks ago
Chew-Seong Cheong
Dec 16, 2020

By Vieta's formula , we have x 1 + x 2 + x 3 = 24 4 = 6 x_1+x_2+x_3 = \dfrac {24}4 = 6 , x 1 x 2 + x 2 x 3 + x 3 x 1 = 44 4 = 11 x_1x_2 + x_2 x_3 + x_3x_1 = \dfrac {44}4 = 11 , and x 1 x 2 x 3 = 23 4 x_1x_2x_3 = \dfrac {23}4 .

Now let a = x 1 x 2 a=x_1x_2 , b = x 2 x 3 b=x_2x_3 , and c = x 3 x 1 c=x_3x_1 and P n = a n + b n + c n P_n = a^n + b^n + c^n , where n n is a positive integer. Then we can solve for P 4 = a 4 + b 4 + c 4 = ( x 1 x 2 ) 4 + ( x 2 x 3 ) 4 + ( x 3 x 1 ) 4 P_4 = a^4 + b^4 + c^4 = (x_1x_2)^4 +(x_2x_3)^4+(x_3x_1)^4 using Newton's sums (or identities) as follows:

Let { S 1 = a + b + c = x 1 x 2 + x 2 x 3 + x 3 x 1 = 11 S 2 = a b + b c + c a = x 1 x 2 2 x 3 + x 1 x 2 x 3 2 + x 1 2 x 2 x 3 = x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) = 23 4 × 6 = 69 2 S 3 = a b c = ( x 1 x 2 x 3 ) 2 = ( 23 4 ) 2 = 529 16 \begin{cases} S_1 = a + b + c = x_1x_2 + x_2 x_3 + x_3x_1 = 11 \\ S_2 = ab+bc+ca = x_1x_2^2x_3 + x_1x_2 x_3^2 + x_1^2x_2x_3 = x_1x_2x_3(x_1+x_2+x_3) = \dfrac {23}4 \times 6 = \dfrac {69}2 \\ S_3 = abc = (x_1x_2x_3)^2 = \left(\dfrac {23}4 \right)^2 = \dfrac {529}{16} \end{cases}

Then

P 1 = S 1 = 11 P 2 = S 1 P 1 − 2 S 2 = 11 × 11 − 2 × 69 2 = 52 P 3 = S 1 P 2 − S 2 P 1 + 3 S 3 = 11 × 52 − 69 2 × 11 + 3 × 529 16 = 4667 16 P 3 = S 1 P 3 − S 2 P 2 + S 3 P 1 = 11 × 4667 16 − 69 2 × 52 + 529 16 × 11 = 7113 4 \begin{aligned} P_1 & = S_1 = 11 \\ P_2 & = S_1P_1 - 2S_2 = 11 \times 11 - 2 \times \frac {69}2 = 52 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 11 \times 52 - \frac {69}2 \times 11 + 3 \times \frac {529}{16} = \frac {4667}{16} \\ P_3 & = S_1P_3 - S_2P_2 + S_3P_1 = 11 \times \frac {4667}{16} - \frac {69}2 \times 52 + \frac {529}{16} \times 11 = \frac {7113}4 \end{aligned}

Therefore p + q = 7113 + 4 = 7117 p+q = 7113+4 = \boxed{7117} .

2 Helpful 1 Interesting 2 Brilliant 0 Confused

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