Mission Accomplished! Topic: Algebra. Level: AIME.

Problem: (1991 India National Olympiad - #7) Solve the following system for real $ x, y, z $

$ x+y-z = 4 $

$ x^2-y^2+z^2 = -4 $

$ xyz = 6 $.

Solution: Well, we simply go about solving this system the regular way - look for something to eliminate. We try using the first and second equations and find

$ (x-z)^2 = (4-y)^2 \Rightarrow x^2-2xz+z^2 = 16-8y+y^2 $,

which we subtract from the second equation to get

$ 2xz = 8y-20 $.

Using the third equation, we have $ 2xz = \frac{12}{y} $ so we substitute, multiply through by $ y $ (it can't be zero), and divide by $ 4 $ to get

$ 0 = 2y^2-5y-3 = (2y+1)(y-3) $.

So $ y = -\frac{1}{2}, 3 $. But from the second equation, we see that $ y = -\frac{1}{2} $ gives $ x^2+z^2 < 0 $ but they are real so this is impossible. Hence $ y = 3 $. It remains to solve for $ x $ and $ z $ using

$ x-z = 1 $

$ xz = 2 $.

Substitute $ z = \frac{2}{x} $ into the first one and again multiply through by $ x $ to find

$ x^2-x-2 = (x-2)(x+1) = 0 $.

We then have $ x = 2, -1 $ with corresponding $ z = 1, -2 $. Checking, we see that both solutions work, so our final solution set is

$ (2, 3, 1) $; $ (-1, 3, -2) $.

QED.

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Comment: Not bad for an olympiad question - just requires basic algebraic manipulation and solving quadratics. Looking for the right things to square and substitute made this problem considerably easier.

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Practice Problem: (1994 India National Olympiad - #2) If $ x^5-x^3+x = a $, prove that $ x^6 \ge 2a-1 $.