A Splashy Math Problem

A problem from the 2021 Moscow Math Olympiad went viral on Russian math channels. The author is Dmitry Krekov.

Problem. 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?



  1. Carl Feynman:


  2. Matthew Rose:


  3. TRidgway:

    Arm wrestle!

  4. Iseng Belajar:

    Someone please provide the proof?

  5. rosie:

    A construction like that of Mills’s constant should do it.

  6. passerby:

    Fun problem! (Solution provided for other visitors I’m sure you had no problem doing this.)

    One way to produce algebraic integers P whose powers are quite close to integers is to observe that the sums of the powers of the conjugates of P are all integral. In particular, if all the other conjugates of P have absolute value less than one, then the closest integer to P^n (for large n) satisfies a linear recurrence relation. For example, if you take P = (sqrt(5)+1)/2 to be the golden ratio, and Q = (-sqrt(5)+1)/2 to be the conjugate of P, then

    PQ = -1, |P| = 1.61803… |Q| = 0.61803…

    And now the sequence P^n + Q^n consists entirely of integers: 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, and so on. These are the Lucas numbers L_n. Now let’s see what happens if we square both sides. We get

    P^(2n) + 2 P^n Q^n + Q^(2n) = 1,9,16,49,… = (L_n)^2.

    But as observed, PQ = -1 so P^n Q^n = (-1)^n, and Q^(2n) is a (positive!) real number less than one. Hence

    P^(2n) + 2 (-1)^n + real number less than one = (L_n)^2.

    But that means the ceiling of P^(2n) differs from (L_n)^2 by 2 or -2, and so you can take A = P^2 in the problem. (If you want the sign of the “2” to be the same, you can take A = P^4 instead.)

    This argument works equally well if P is any unit in a real quadratic number field, and A=P^2. So you could also take A = (1 + sqrt(2))^2 = (3+2 sqrt(2)) or (2 + sqrt(3))^2 = (7 + 2 sqrt(3)), etc.

  7. carl feynman:

    You are right and I am wrong! Good puzzle.

  8. rosie:

    “the sums of the powers of the conjugates of P are all integral” Whoa — plural overload! Which conjugates? which powers? What are the terms you’re summing? and for each sum what is the set of terms for that sum?

    I think I know a bit about algebraic conjugates but I didn’t understand what I found. How do you find the algebraic conjugates of a number? It seemed to me that the process was defined, not on a number itself, but only on the representation of the number in a certain form, and it gave you another representation.

  9. Tanya Khovanova's Math Blog » Blog Archive » A Splashy Math Problem Solution:

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