# Interesting Algebra Problem

• Simple Induction
• PeroK

#### PeroK

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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:
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.