The Red Capes are coming!

Algebra Level 5

$\begin{cases} a+b+c= 0 \\ a^5 + b^5 + c^5 = 5 \\ a^6 + b^6 + c^6 = 6 \end{cases}$

Let $a,b$ and $c$ be numbers satisfying the system of equations above. And at least one of $a,b$ and $c$ is not a real number.

Given that $a^2+b^2+c^2$ satisfy a least degree monic polynomial $f(X)$ , find $f(2)$ .

The answer is -16.

1 solution

Alan Enrique Ontiveros Salazar
May 6, 2016

Let's use the standard notation of Newton's Sums, i.e., $P_n=a^n+b^n+c^n$ , $S_1=a+b+c$ , $S_2=ab+ac+bc$ and $S_3=abc$ . Then we have:

$P_1=S_1=0 \implies S_1=0$

$P_2=S_1P_1-2S_2 \implies P_2=-2S_2$

$P_3=S_1P_2-S_2P_1+3S_3 \implies P_3=3S_3$

$P_4=S_1P_3-S_2P_2+S_3P_1 \implies P_4=2S_2^2$

$P_5=S_1P_4-S_2P_3+S_3P_2=5 \implies -5S_2S_3=5 \implies S_3=-\dfrac{1}{S_2}$

$P_6=S_1P_5-S_2P_4+S_3P_3=6 \implies -2S_2^3+3S_3^2=6 \implies -2S_2^3+3\left(-\dfrac{1}{S_2}\right)^2=6$

So, the minimal polynomial of $S_2$ is $P(x)=2x^5+6x^2-3$ . Substituting $S_2=-\dfrac{P_2}{2}$ we get the minimal monic polynomial of $P_2$ : $f(x)=x^5-24x^2+48$ (it is irreducible by Eisentein's criterion for $p=3$ ). Then $f(2)=\boxed{-16}$ .

You need to prove that it the polynomial is minimal and at least one of $a,b$ and $c$ is not real.

Pi Han Goh - 5 years, 1 month ago

Done, but I don't see the need to prove if $a,b$ or $c$ are real or not.

Alan Enrique Ontiveros Salazar - 5 years, 1 month ago

To prove that there exists $a,b,c$ such that it satisfy these conditions.

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – We need that the discriminant of the polynomial whose roots are $a,b,c$ , i.e. $Q(x)=x^3-S_1x^2+S_2x-S_3$ , be less than zero. That is, $\Delta=-4S_2^3-27S_3^2$ . But $-2S_2^3+3S_3^2=6$ , so we need to prove that there is at least one $S_2$ such that $S_2^3>-\dfrac{27}{11}$ or $S_2>-1.34893$ . We see that $P(0)<0$ and $P(1)>0$ , so we have one $S_2$ such that $0<S_2<1$ , which satisfies our inequality.

Alan Enrique Ontiveros Salazar - 5 years, 1 month ago

@Alan Enrique Ontiveros Salazar – For your $-2S_2^2 + 3S_2 ^2 = 6$ , use descartes' rule of signs followed by intermediate value theorem and you're done.

Great job by the way. I didn't expect such a quick solution to appear! + 1

Plus, long time didn't see u on this site!

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Your argument seems to be assuming that $S_2$ and $S_3$ must be real...

Any root $\alpha$ of $2x^5 - 6x^2 + 3 = 0$ yields a cubic equation $x^3 - \alpha x - \alpha^{-1} = 0$ whose roots are an acceptable set of numbers $a,b,c$ . If $\alpha$ is not real, the cubic equation can only have at most one real root (the unique real number $-1/|\alpha|^2\,$ that makes the imaginary part of the cubic expression vanish is the only possible candidate), and so must have at least one complex root.

Since the quintic has three real roots and two complex roots, we are done (by choosing a complex root $\alpha$ ).

Mark Hennings - 5 years, 1 month ago

@Mark Hennings – Wait what? The sum of at least 3 distinct powers of $a,b,c$ are real, so by Newton's identities, the (least degree) polynomial that satisfy $a,b,c$ must have all real coefficients. No? Or is my reasoning just hocus pocus?

EDIT: How did you convert "any root $\alpha$ of $2x^5-6x^2+3=0$ yields a cubic equation $x^3-\alpha x - \alpha^{-1} = 0$ whose roots are an acceptable set of numbers $a,b,c$ " in the first place? Does this got to do with Galois Theory? because I'm still only in the "Background" chapter.... Or is it some depressing the cubic /Cardanos' method deal?

EDIT2: It's like you're implying that there exists non-real $S_2$ and $S_3$ such that the roots of $a,b,c$ satisfy the cubic equation $x^3 + S_2 x - S_3 = 0$ and satisfy the 3 given equations.

EDIT3: I'll come back to your comment later.

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Suppose that $a,b,c$ satisfy the cubic $x^3 - \alpha x - \beta = 0$ . Then $P_5 = 5\alpha\beta$ and $P_6 = 2\alpha^3 + 3\beta^2$ . The conditions that $P_5=5$ and $P_6=6$ are satisfied provided that $\alpha\beta=1$ and $2\alpha^5 - 6\alpha^2 + 3 = 0$ . Thus any solution of the quintic yields an acceptable set of $a,b,c$ .

Since the quintic has 2 real turning points, it has (at most) three real roots, and hence (at least) two complex roots. In fact, it has exactly two complex roots.

One of the complex solutions of the quintic is (approximately) $\alpha = -0.68496 - 1.34346 i$ . The cubic equation it generates has roots $-0.815917 + 0.908841 i$ , $0.257173 + 0.31206 i$ and $0.558744 - 1.2209 i$ , which do indeed have sums of their fifth and sixth powers equal to $5$ and $6$ respectively.

Mark Hennings - 5 years, 1 month ago

@Mark Hennings – OHHHHH NOOOOOOOO! my question has been demolished! Let me see how to amend it...

On an unrelated topic, Galois theory is super duper interesting. I got to see how we can't trisect 60 degrees using normal constructions, and how we can show that there is no solution in radicals for some 5th degree polynomial.... super cooooooooool! Thanks for introducing it to me!!

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – It is not demolished. The answer is still correct, but there are just more ways of solving it than you thought! You have shown that there exists a real $\alpha$ which gives a solution, and I have shown that there are two complex $\alpha$ that give a solution. That is all.

Mark Hennings - 5 years, 1 month ago

@Mark Hennings – Oh right! Got it! I'm under the impression that if you made a comment on something, then that something must be very wrong. Hahaha!

For what it's worth, here's how I've done it (at least in a clearer manner):

$S_2$ satisfy the equation $2X^5 + 6X^2 - 3 = 0$ . By Eisenstein's irreducibility criterion , this polynomial is irreducible when $p=3$ .

Let $f(X) = 2X^5 + 6X^2 - 3$ .
Then $f(X)$ has 1 change of signs in the polynomial's coefficients, and
$f(-X)$ has 2 change of signs in the polynomial's coefficients.

By Descartes' rule of signs , there is at most 3 real roots of $f(X)$ , or equivalently, there is either 2 or 4 complex roots of $S_2$ , and we're done!!!

Thanks anyway!!

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Thanks, also nice problem :D

Alan Enrique Ontiveros Salazar - 5 years, 1 month ago

@Alan Enrique Ontiveros Salazar – Hhaaha! This is a nicest problem

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Now I'm trying to solve this .

Alan Enrique Ontiveros Salazar - 5 years, 1 month ago

@Alan Enrique Ontiveros Salazar – Hint: what is everyone's favorite answer on this site?

Pi Han Goh - 5 years, 1 month ago

The Red Capes are coming!

The answer is -16.

1 solution

0 pending reports