Find the smallest real number $C$, such that for any positive integers $x \neq y$ holds the following: $$\min(\{\sqrt{x^2 + 2y}\}, \{\sqrt{y^2 + 2x}\})<C$$ Here $\{x\}$ denotes the fractional part of $x$. For example, $\{3.14\} = 0.14$. Proposed by Anton Trygub
Problem
Source: Ukrainian Mathematical Olympiad 2023. Day 1, Problem 9.4
Tags: algebra, fractional part
Tintarn
05.04.2023 10:02
Nice problem!
The best possible constant is $\varphi-1=\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2} \approx 0.618$.
Indeed, a construction with large $x$ and $y$ satisfying $y =\lfloor \varphi x\rfloor$ (say) has $x \ne y$ and
\[\sqrt{x^2+2y}=\sqrt{x^2+2\varphi x+\mathcal{O}(1)}=x+\varphi+o(1)\]and
\[\sqrt{y^2+2x}=\sqrt{y^2+\frac{2y}{\varphi}+\mathcal{O}(1)}=y+\frac{1}{\varphi}+o(1)\]and since $\{\varphi\}=\left\{\frac{1}{\varphi}\right\}=\varphi-1$, this constitutes a counterexample for any $C>\varphi-1$ and $x,y$ large.
On the other hand, let us show that $C=\varphi-1$ works.
For the sake of contradiction, suppose that $x \ne y$ satisfies $\{\sqrt{x^2+2y}\} \ge C$ and $\{\sqrt{y^2+2x}\} \ge C$ and w.l.o.g. $x<y$.
Then the second condition implies $\sqrt{y^2+2x} \ge y+C$ and hence $2x \ge 2Cy+C^2$ i.e. $x>Cy$.
Since $y>x$, we have $x^2+2y>x+1$ and hence the first condition implies $\sqrt{x^2+2y} \ge x+C+1$ and hence $y>(C+1)x$.
But $C+1=\frac{1}{C}$, so these two inequalities contradict each other for $C=\varphi-1$. Done!
v_Enhance
15.04.2023 05:10
Solution from Twitch Solves ISL: The answer is $C = \frac{\sqrt5-1}{2}$. First, we prove this works. WLOG, $x < y$. Then observe two cases: Whenever $x < Cy$, we have \[ y < \sqrt{y^2+2x} < y + C. \] Also, whenever $Cy < x < y$ we have \[ x+1 < \sqrt{x^2+2y} < x + (C+1). \] This, in each case of the fractional parts is less than $C$. Conversely, for the construction, we really just need $x/y \approx C$. Consider any increasing sequences $x_n$, $y_n$ of positive integers with $\lim_{n \to \infty} \frac{x_n}{y_n} = C$. (For example, one could take $y_n = 10^n$ and $x_n = \left\lfloor 10^n C \right\rfloor$, or take $y_n$ and $x_n$ to be consecutive Lucas/Fibonacci numbers.) For brevity, abbreviate $x_n$ to $x$ and $y_n$ to $y$. Then \begin{align*} \sqrt{y^2+2x} - y &= \frac{2x}{\sqrt{y^2+2x}+y} = \frac{2}{\sqrt{\left( \frac yx \right)^2 + \frac 2x} + \frac yx} \\ &\xrightarrow{n \to \infty} \frac{2}{\sqrt{(1/C)^2}+(1/C)} = C. \end{align*}and similarly \begin{align*} \sqrt{x^2+2y} - (x+1) &= \frac{2y-2x-1}{\sqrt{x^2+2y}+x+1} = \frac{2 - \frac{1}{y-x}} {\sqrt{\left( \frac x{y-x} \right)^2 + \frac2{y-x}} + \frac x{y-x} + \frac{1}{y-x}} \\ &\xrightarrow{n \to \infty} \frac{2}{\sqrt{\left( \frac{C}{1-C} \right)^2} + \frac{C}{1-C}} = C. \end{align*}
emi3.141592
13.11.2024 19:57
The answer is $c = \frac{1}{\phi}$. To show that this constant works, we consider the following cases: Case 1. If $\sqrt{x^2 + 2y} \geq x + 2$, then $y \geq 2x + 2$. It follows that $$ \{\sqrt{y^2+2}\}= \frac{2x}{\sqrt{x^2 + 2x} + y} \leq \frac{2x}{\sqrt{x^2 + 10x + 4} + 2x + 2} \leq \frac{2}{5} < \frac{1}{\phi}. $$ Case 2. If $\sqrt{x^2 + 2x} < x + 2$, suppose there exist $y \leq x$ such that $$ \sqrt{y^2 + 2x} \geq y + c \quad \text{and} \quad \sqrt{x^2 + 2x} \geq (x + 1) + c. $$We observe that $$ \sqrt{y^2 + 2x} \geq y + c \Rightarrow 2x \geq 2yc + c^2 \Rightarrow x > yc. $$Similarly, we have that $$ \sqrt{x^2 + 2x} \leq x + (c + 1) \Rightarrow 2x \leq 2x(c + 1) + (c + 1)^2 \Rightarrow x \leq x(c + 1). $$The last two inequalities imply that $\frac{1}{c} > \frac{y}{x} > c + 1$, which is not possible, as $\frac{1}{\phi} = \phi + 1$, and we are done.