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?
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
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.
“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.
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 amcarl feynman:
You are right and I am wrong! Good puzzle.
27 June 2021, 10:09 amrosie:
“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.
22 July 2021, 9:28 amTanya Khovanova's Math Blog » Blog Archive » A Splashy Math Problem Solution:
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