Where's the First Powers?

Algebra Level 5

Given that the complex numbers x , y , z x,y,z satisfy the equations

x 2 + y 2 + z 2 = 2 x 3 + y 3 + z 3 = 3 x 4 + y 4 + z 4 = 4 \begin{aligned} x^2 + y^2 + z^2 & = & 2 \\ x^3 + y^3 + z^3 & = & 3 \\ x^4 + y^4 + z^4 & = & 4 \\ \end{aligned}

Denote f f as a minimum degree monic polynomial with real coefficients such that it has root x + y + z x+y+z .

What is the value of f ( 1 ) f(1) ?


The answer is 1.

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Pi Han Goh
Apr 30, 2014

Let S n S_n denote the n th n^{\text{th}} symmetric sum for positive integer n n . We have

( x + y + z ) 2 = ( x 2 + y 2 + z 2 ) + 2 ( x y + x z + y z ) ⇒ S 1 2 − 2 S 2 = 2 (x+y+z)^2 = (x^2 + y^2 + z^2) + 2(xy+xz+yz) \Rightarrow S_1^2 - 2S_2 = 2

With the algebraic identity

x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( ( x + y + z ) 2 − 3 ( x y + x z + y z ) ) x^3 + y^3 + z^3 - 3xyz = (x+y+z) \left ( (x+y+z)^2 - 3(xy+xz+yz) \right )

⇒ 3 S 3 = 3 − S 1 ( S 1 2 − 3 S 2 ) = 3 − S 1 ( 2 − S 2 ) ⇒ S 3 = 3 − S 1 ( 2 − S 2 ) 3 \Rightarrow 3S_3 = 3 - S_1 ( S_1^2 - 3S_2 ) = 3 - S_1 (2 - S_2) \Rightarrow S_3 = \frac {3 - S_1 (2 - S_2) }{3 }

And

( x 2 + y 2 + z 2 ) 2 = ( x 4 + y 4 + z 4 ) + 2 ( ( x y ) 2 + ( x z ) 2 + ( y z ) 2 ) (x^2 + y^2 + z^2)^2 = (x^4+y^4+z^4) + 2 \left ( (xy)^2 + (xz)^2 + (yz)^2 \right )

Because ( x y ) 2 + ( x y ) 2 + ( x z ) 2 = ( x y + x z + y z ) 2 − 2 x y z ( x + y + z ) (xy)^2 + (xy)^2 + (xz)^2 = (xy+xz+yz)^2 - 2xyz(x+y+z) , then S 2 2 = 2 S 1 S 3 S_2^2 = 2S_1 S_3

Substitution of S 3 S_3 and followed by substitution of S 2 S_2

S 2 2 = 2 S 1 3 ⋅ ( 3 − S 1 ( 2 − S 2 ) ) ( S 1 2 − 2 2 ) 2 = 2 S 1 3 ⋅ ( S 1 2 − 2 S 1 + 1 ) S 1 4 − 12 S 1 2 + 24 S 1 − 12 = 0 \begin{aligned} S_2^2 & = & \frac {2S_1}{3} \cdot (3 - S_1 (2 - S_2 ) ) \\ \left ( \frac { S_1^2 - 2}{2} \right )^2 & = & \frac {2 S_1}{3} \cdot (S_1^2 - 2S_1 + 1 ) \\ S_1^4 - 12S_1^2 + 24S_1 - 12 & = & 0 \\ \end{aligned}

By Eisenstein's criterion . With p = 3 p = 3 , w 4 − 12 w 2 + 24 w − 12 w^4 - 12w^2 +24w-12 is indeed irreducible over the rational numbers. Hence, it is the minimal polynomial

So f ( w ) = w 4 − 12 w 2 + 24 w − 12 ⇒ f ( 1 ) = 1 f(w) = w^4 - 12w^2 + 24w - 12 \Rightarrow f(1) = \boxed{1}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Well i am not afraid to say but my answer was correct by chance

Parth Lohomi - 6 years, 9 months ago

Ideally, you should show that the polynomial is minimal, ie that it cannot be factored further.

Nice use of Newtons equations.

Calvin Lin Staff - 7 years, 1 month ago

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Addendum:

By Eisenstein's criterion . With p = 3 p = 3 , w 4 − 12 w 2 + 24 w − 12 w^4 - 12w^2 +24w-12 is indeed irreducible over the rational numbers. So f ( w ) = w 4 − 12 w 2 + 24 w − 12 f(w) = w^4 - 12w^2+24w-12

Thanks!

Pi Han Goh - 7 years, 1 month ago

I could not get How you wrote S1^{2}-2S2=2

Vaibhav Agrawal - 7 years, 1 month ago

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My bad, I defined S n S_n wrongly, I meant to say that S n S_n is the n-th elmentary symmetric sum . Fixed!

Pi Han Goh - 7 years, 1 month ago

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