What is \varphi?

$$\frac{1+\sqrt{5}}{2}$$

25th Olympic Revenge wrote: Prove that there exists a positive integer $x<5^{2022}$ such that \[\{\varphi\sqrt[3]{x}\}<\varphi^{-2022}.\] Solved with Pujnk Consider $x = \lfloor \varphi^{3 \cdot 2022} \rfloor$. Note that $x < (\varphi^3)^{2022} < 5^{2022}$. We'll start by proving some basic bounds: Claim 01. $\lfloor \varphi^{2023} \rfloor = \varphi^{2023} - \frac{1}{\varphi^{2023}} $ Proof. Indeed, note that $\varphi$ and $-\frac{1}{\varphi}$ are roots of $x^2 - x - 1 = 0$. Therefore, we have $\varphi^n + \left( - \frac{1}{\varphi} \right)^n \in \mathbb{Z}$ for all $n \in \mathbb{N}_0$. This implies $\varphi^{2023} - \frac{1}{\varphi^{2023}} \in \mathbb{Z}$. Now, note that \[ \varphi^{2023} - \left( \varphi^{2023} - \frac{1}{\varphi^{2023}} \right) = \frac{1}{\varphi^{2023}} < 1 \implies \lfloor \varphi^{2023} \rfloor = \varphi^{2023} - \frac{1}{\varphi^{2023}} \] Claim 02. $\lfloor \varphi^{2023} \rfloor \le \lfloor \varphi \sqrt[3]{x} \rfloor$ Proof. We first claim $\varphi \sqrt[3]{x} > \varphi^{2023} - \varphi^{-2023}$. Indeed, \begin{align*} \varphi \sqrt[3]{x} &> \varphi \sqrt[3]{\varphi^{3 \cdot 2022} - 1} \\ &= \sqrt[3]{\varphi^{3 \cdot 2023 } - \varphi^3} \\ &> \sqrt[3]{\varphi^{3 \cdot 2023} - 3 \cdot (\varphi^{2023} - \varphi^{-2023}) - \varphi^{-(3 \cdot 2023)}} = \varphi^{2023} - \varphi^{-2023} \end{align*}This therefore implies $\lfloor \varphi \sqrt[3]{x} \rfloor \ge \varphi^{2023} - \varphi^{-2023} = \lfloor \varphi^{2023} \rfloor$. We are now ready to tackle the problem. \begin{align*} \{ \varphi \sqrt[3]{x} \} &= \varphi \sqrt[3]{x} - \lfloor \varphi \sqrt[3]{x} \rfloor \\ &\le \varphi^{2023} - \lfloor \varphi^{2023} \rfloor = \varphi^{-2023} < \varphi^{-2022} \end{align*}as desired.

adic wrote: What is \varphi? \var stands for "variant." So, \varphi refers to a variant symbol of phi: $\varphi$, while \phi refers to the generally written symbol of phi: $\phi$.

Note that $\varphi\sqrt[3]{x+1}-\varphi\sqrt[3]{x}=\frac{\varphi}{\sqrt[3]{x^2}+\sqrt[3]{x(x+1)}+\sqrt[3]{(x+1)^2}}<\frac{\varphi}{x^{\frac{2}{3}}}$. So $\varphi\sqrt[3]{x+1}-\varphi\sqrt[3]{x}<\varphi^{-2022}$ for all $x>\varphi^{3036}$. Given that $5>\varphi^3$, we have $5^{1014}>\varphi^{3036}$ and $\varphi\sqrt[3]{5^{2019}}-\varphi\sqrt[3]{5^{1014}}=\varphi(5^{673}-5^{338})>1$. So $\lfloor \varphi\sqrt[3]{x}\rfloor$ must increase for some integer $x\in(\varphi^{3036}, 5^{2022})$. Taking this $x$, $\{\varphi\sqrt[3]{x}\}=\varphi\sqrt[3]{x}-\lfloor \varphi\sqrt[3]{x}\rfloor<\varphi\sqrt[3]{x}-\varphi\sqrt[3]{x-1}<\varphi^{-2022}$.