﻿ Another Youtube Problem | Observations

Recently, I found the following problem on Youtube:

$$x+y+z=1, \\ x^2+y^2+z^2=2, \\ x^3+y^3+z^3=3, \\ x^n+y^n+z^n= \,?$$

I did not continue to watch the video. So I have no idea what solution is presented. I decided to treat that as a challenge and to observe my steps, including the failed ones.

The problem sounds like a normal set of three equations with three unknowns that we can solve by eliminating. After all, this is the standard technique that we learn in school. But it is actually not that easy. Eliminating x from the second equation and solving for y yields an expression in y which contains a square root. Inserting everything in the third equation results in a mess.

Let us try numerically. We use Euler Math Toolbox here.

>function f2(x,y) &= 2-x^2-y^2-(1-x-y)^2

2                2    2
- y  - (- y - x + 1)  - x  + 2

>a3 &:= 3
3
>function f3(x,y) &= a3-x^3-y^3-(1-x-y)^3

3                3    3
- y  - (- y - x + 1)  - x  + 3

>plot2d("f2",levels=0,r=2); We see, that the equations do not have a real solution. But they have, of course, a complex solution by the main theorem of Algebra.

The second idea is to try some tricks on the equations. E.g., we could multiply the first equation with x, y, and z and add the three results. Because of the first and the second equation, we get

$$xy+yz+xz=-\dfrac{1}{2}$$

That is a new equation. But is not clear how this is going to help.

After a while, the mathematical instinct kicks in to try something more generic and simple. Pattern matching inside the brain brings a known trick to the mind. Any sequence

$$1,x,x^2,x^3,\ldots$$

has an interesting property. If x is the zero of a polynomial

$$p(t)=t^3+at^2+bt+c$$

It satisfies the interesting property

$$x^n+ax^{n-1}+bx^{n-2}+cx^{n-3}=0$$

Thus the sequence above satisfies a recursion formula. Now, if x,y,z are the three zeros of the polynomial p(t) all three satisfy the same recursion formula, and so does the sum of the three zeros.

$$s_n = – (as_{n-1}+bs_{n-2}+cs_{n-3})$$

for

$$s_n = x^n+y^n+z^n$$

So we know from our three equations and the trivial equation

$$s_0 = x^0+y^0+z^0=3$$

that

$$3 = -(2a+b+3c)$$

But we also know that

$$p(t) = (t-x)(t-y)(t-z) = t^3 – (x+y+z) t^2 + (xy+yz+xz)t -xyz$$

Once we see that, we get a=-1 immediately from our problem. Having played around with tricks, we also know b=-1/2. From that, we now have

$$a=-1, \quad b=-\dfrac{1}{2}, \quad c=-\dfrac{1}{6}$$

We can verify that with Euler Math Toolbox.

For that, we compute the zeros of the polynomial and check the equations numerically. As expected, two of the zeros are complex.

>s=polysolve([-1/6,-1/2,-1,1])
[ -0.215425+0.264713i,  -0.215425-0.264713i,  1.43085+0i  ]
>sum(s)
1+0i
>sum(s^2)
2+0i
>sum(s^3)
3+0i


The expression for the real zero of p(t) is terrible and involves a third root.

$$z=u^{1/3} + \dfrac{5}{u^{1/3}} + \dfrac{1}{3}, \quad u=\dfrac{\sqrt{13}}{9 \cdot 2^{2/3}}+\dfrac{11}{54}$$

Thus, we get a recursion for the sum and thus for the equation sign in our problem.

$$s_n = s_{n-1}+\dfrac{1}{2}s_{n-2}+\dfrac{1}{6}s_{n-3}$$

Check with EMT.

>fraction sequence("x[-1]+x[-2]/2+x[-3]/6",[3,1,2,3],10)
[3,  1,  2,  3,  25/6,  6,  103/12,  221/18,  1265/72,  905/36]
>fraction real(sum(s^[0:9]')')
[3,  1,  2,  3,  25/6,  6,  103/12,  221/18,  1265/72,  905/36]


Can we make a problem with an easier solution? Sure! We only have to start with simpler zeros.

$$x+y+z=0, \\ x^2+y^2+z^2=14, \\ x^3+y^3+z^3=-18, \\ x^n+y^n+z^n= \,?$$

$$s_n = 1 + 2^n + (-3)^n$$