Alexander Karabegov’s Puzzle
When I was in 8th grade, I was selected to be part of the Moscow math team and went to Yerevan, Armenia, to participate in the All-Soviet Math Olympiad. A group of us boarded a bus, and Alexander Karabegov paid for all of our bus tickets. He was from Yerevan himself and wanted to be a gracious host. I was impressed. The next time I met him was when I started studying at the Moscow State University. We have been friends ever since. He was even the best man at one of my weddings. Now, he lives in Texas and sends me his original puzzles from time to time. Today, he sent me a new one.
WARNING. His solution to the puzzle is also included. So if you want to solve it yourself, stop reading after the next paragraph.
Puzzle. A number c is called a fixed point of a function f, if it is a solution of the equation f(x) = x; that is, if f(c) = c. Find all solutions of the equation g(g(x)) = x, where g(x) = x2 + 2x − 1; that is, find all fixed points of the function f(x) = g(g(x)). (We can assume that x is a real number.)
I gave the puzzle to my students, and they converted it to a fourth-order equation, which they solved using various methods. What I liked about Alexander’s solution is it only uses quadratic equations. I am too lazy to give his full solution. Here is just his solve path.
Solve path. If c is a fixed point of the function g(x), then it is a fixed point of f(x) = g(g(x)). Solving the equation g(c) = c gives us two fixed points. We need two more, as our equation is quartic. Suppose a is another fixed point. Let b = g(a). It follows that g(b) = a. Moreover, we can assume that a is not b, as we covered this case before. We get two equations a2 + 2a − 1 = b and b2 + 2b − 1 = a. Subtracting one equation from another, we get a quadratic equation that has to be divisible by a −b. As b is not a, by our assumption, we can divide the result by a − b, expressing b as a linear function of a. We plug this back into one of the two equations and get a quadratic equation for a, supplying us with the remaining two solutions. TADA!
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Zarunias:
My path was this: I just substitute g(x) with x²+2x-1 into the equation g(g(x))=x twice, to get x^4+4x³+4x²-x-2=0. As you mentioned a fixed point of the function g(x) is a solution to the equation, so I already know two solutions, which are the solutions of x²+x-1=0.
So I know that I can factor the quartic into this quadratic and another quadratic, this gets us to x^4+4x³+4x²-x-2=(x²+x-1)(x²+3x+2)=0, and from this it’s easy to get all four solutions.
Yes, there is a quartic in the solution, but I want to argue that from the beginning on it’s clear that I can divide it by the two known solutions, giving us a quadratic. In the end it’s very similar to Alexander’s solution except that I don’t introduce new variables.
2 November 2024, 9:33 am