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 number A so that for any natural number n, there exists a square of a natural number that differs from the ceiling of An by 2?
The problem is very difficult, but the solution is not long. It starts with a trick. Suppose A = t2, then An + 1/An = t2n + 1/t2n = (tn + 1/tn)2 − 2. If t < 1, then the ceiling of An differs by 2 from a square as long as tn + 1/tn 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 x2 – 6x + 1, and declare t to be its largest root.Share: