6. [10] Find at least one real number $A$ with the property: for any natural number $n$, the distance from the ceiling of the number $A^{n}$ to the nearest square of an integer is 2. (The ceiling of a number $x$ is the smallest integer not less than $x$.)

Dmitry Krykov
Solution. Consider any quadratic equation with integer coefficients and the leading coefficient 1, which has two positive roots, the product of which is 1. For example, the equation $x^{2}-4 x+1$ has roots $2+\sqrt{3}$ and $2-\sqrt{3}$. Note that the sum and product of these roots are integers, and then the sum $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is an integer for any natural $n$ (this is not difficult to prove by induction or simply by expanding the brackets: terms with $\sqrt{3}$ either enter in an even power or cancel each other out).

Then $\left((2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right)^{2}-$ is a perfect square, and it equals $(2+\sqrt{3})^{2 n}+2+(2-\sqrt{3})^{2 n}$ (since the product of the roots is 1), which is 2 away from the number $(2+\sqrt{3})^{2 n}+(2-\sqrt{3})^{2 n}$, which, in turn, is the upper integer part of the number $(2+\sqrt{3})^{2 n}$ (since the second root is positive and less than 1).

But then the number $A=(2+\sqrt{3})^{2}$ is the desired one.

Comment. It is not difficult to see that for $t$ we can take any number that is the larger root of a polynomial of the form $x^{2}-n x+1=0$, where $n$ is a natural number not less than 3. Indeed, as in the solution above, the sum of the roots $t^{n}+\frac{1}{t^{n}}$ of this polynomial turns out to be an integer, from which for $A=t^{2}$ the statement of the problem follows.

In this solution, we see that for the numbers we have taken, the distance from the power $t^{n}$ to the nearest integer tends to zero as $t$ increases. In fact, there are more numbers whose powers become closer and closer to integers (but we cannot say that they fit for solving this problem!).

Namely, let $P(x)$ be a monic polynomial with integer coefficients, all of whose roots (including complex ones), except one, are less than 1 in absolute value. Then this root $x_{1}$ is real, and the distance from $x_{1}^{n}$ to the nearest integer tends to 0 as $n$ increases. This follows from the fact that the sum of the $n$-th powers of all the roots of the polynomial $P(x)$ is expressed as an integer through its coefficients, and is therefore an integer. And the powers of all other roots tend to 0 - precisely because they are less than 1 in absolute value. This reasoning can be found in the article by A. Egorov “Pisot Numbers” (journal “Kvant”, issues 5 and 6 for 2005); see also the project “Fractional Parts of Powers” at the XII Summer Conference of the Tournament of Towns.

Such numbers - roots of a monic polynomial with integer coefficients, all other roots of which are less than 1 in absolute value - are called Pisot numbers or Pisot-Vijayaraghavan numbers. They are of interest in connection with Diophantine approximation problems and have been studied in the works of Thue, Hardy, Pisot (see, for example, the book: J. W. S. Cassels. Introduction to Diophantine Approximations. Moscow: IL, 1961. [Chapter VIII]).

These numbers received their name after the publication of Charles Pisot, who discovered many remarkable properties of these numbers in his dissertation.