## A Splashy Math Problem Solution

I recently wrote a post, A Splashy Math Problem, with an interesting problem from the 2021 Moscow Math Olympiad.

Problem (by Dmitry Krekov).Does there exist a numberAso that for any natural numbern, there exists a square of a natural number that differs from the ceiling ofA^{n}by 2?

The problem is very difficult, but the solution is not long. It starts with a trick. Suppose *A* = *t*^{2}, then *A*^{n} + 1/*A*^{n} = *t*^{2n} + 1/*t*^{2n} = (*t*^{n} + 1/*t*^{n})^{2} − 2. If t < 1, then the ceiling of *A*^{n} differs by 2 from a square as long as *t*^{n} + 1/*t*^{n} is an integer. A trivial induction shows that it is enough for *t* + 1/*t* to be an integer. What is left to do is to pick a suitable quadratic equation with the first and the last term equal to 1, say *x*^{2} – 6*x* + 1, and declare *t* to be its largest root.

## Roy:

Nice! sounds like a very interesting problem.

29 December 2021, 10:24 am