Interesting Algebra Problem

[PeroK]

I found this interesting video from Presh Talwalkar:




Problem Statement. If:
$$x + y + z = 1$$$$x^2 + y^2 + z^2 = 2$$$$x^3 + y^3 + z^3 = 3$$ Then, find the value of the higher powers such as $$x^5 + y^5 + z^5$$
The solution posted there uses the full Girard-Newton Identities. Here is an elementary solution using the same ideas:

Spoiler

First, note that $$x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + xz + yz)$$ and, for $n \ge 3$ (which is easy to verify): $$x^n + y^n + z^n = (x + y +z)(x^{n-1} + y^{n-1} + z^{n-1}) - (xy + xz + yz)(x^{n-2} + y^{n-2} + z^{n-2}) + xyz(x^{n-3} + y^{n-3} + z^{n-3})$$ Plugging the values we have in the first equation gives:
$$2 = 1 - 2(xy + xz + yz)$$ Hence $$xy + xz + yz = -\frac 1 2$$ Then, for $n = 3$ we have:
$$3 = (1)(2) - (-\frac 1 2)(1) + xyz(3)$$ Giving $$xyz = \frac 1 6$$ Then, for $n = 4$ we have:
$$x^4 + y^4 + z^4 = 3 - (-\frac 1 2)(2) + \frac 1 6 = \frac{25}{6}$$ And, for $n = 5$ we have:
$$x^5 + y^5 + z^5 = \frac{25}{6} - (-\frac 1 2)(3) + (\frac 1 6)(2) = 6$$
In general we have $$x^n + y^n + z^n = (x^{n-1} + y^{n-1} + z^{n-1}) + \frac 1 2(x^{n-2} + y^{n-2} + z^{n-2}) + \frac 1 6 (x^{n-3} + y^{n-3} + z^{n-3})$$ And that, in fact, $x^n + y^n + z^n$ must be rational for all $n$.

Finally, note that we can generalise this procedure for any initial values of the first three expressions.

 

[mfb]

WolframAlpha only finds a complex solution. All the steps are possible with complex numbers, of course.
x = 1.431, y = -0.215-0.265i, z = -0.215+0.265i

 

[PeroK]

WolframAlpha only finds a complex solution. All the steps are possible with complex numbers, of course.
x = 1.431, y = -0.215-0.265i, z = -0.215+0.265i

Yes, I forgot to add a note to say that the underlying solutions for $x, y, z$ are complex. Early in the video he uses Wolfram Alpha as well!