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?
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Carl Feynman:
No.
4 April 2021, 6:35 pmMatthew Rose:
Yes.
19 April 2021, 5:38 amTRidgway:
Arm wrestle!
23 April 2021, 3:45 pmIseng Belajar:
Someone please provide the proof?
29 April 2021, 5:32 amrosie:
A construction like that of Mills’s constant should do it.
16 May 2021, 3:32 ampasserby:
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.
27 May 2021, 11:03 am