The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for every integer $n$. Let $k$ be a fixed integer. A sequence $(a_n)$ of integers is said to be $\textit{phirme}$ if $a_n + a_{n+1} = F_{n+k}$ for all $n \geq 1$. Find all $\textit{phirme}$ sequences in terms of $n$ and $k$.

Let $ABCD$ be a quadrilateral circumscribed around a circle $\omega$ with center $I$. Assume $P$ and $Q$ are distinct points and are isogonal conjugates such that $P, Q$, and $I$ are collinear. Show that $ABCD$ is a kite, that is, it has two disjoint pairs of consecutive equal sides.

Let $n$ be a positive integer. On a blackboard is a circle, and around it $n$ non-negative integers are written. Raphinha plays a game in which an operation consists of erasing a number $a$ neighboring $b$ and $c$, with $b \geq c$, and writing in its place $b + c$ if $b + c \leq 5a/4$ and $b - c$ otherwise. Your goal is to make all the numbers on the board equal $0$. Find all $n$ such that Raphinha always manages to reach her goal.

We say that a prime $p$ is $n$-$\textit{rephinado}$ if $n | p - 1$ and all $1, 2, \ldots , \lfloor \sqrt[\delta]{p}\rfloor$ are $n$-th residuals modulo $p$, where $\delta = \varphi+1$. Are there infinitely many $n$ for which there are infinitely many $n$-$\textit{rephinado}$ primes? Notes: $\varphi =\frac{1+\sqrt{5}}{2}$. We say that an integer $a$ is a $n$-th residue modulo $p$ if there is an integer $x$ such that $$x^n \equiv a \text{ (mod } p\text{)}.$$

Let $ABC$ be a triangle and $H$ and $D$ be the feet of the height and bisector relative to $A$ in $BC$, respectively. Let $E$ be the intersection of the tangent to the circumcircle of $ABC$ by $A$ with $BC$ and $M$ be the midpoint of $AD$. Finally, let $r$ be the line perpendicular to $BC$ that passes through $M$. Show that $r$ is tangent to the circumcircle of $AHE$.

The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for every integer $n$. A sequence $(a_n)$ of integers is said to be $\textit{phirme}$ if there is a fixed integer $k$ such that $a_n + a_{n+1} = F_{n+k}$ for all $n \geq 1$. Show that if $(a_n)$ is a $\textit{phirme}$ sequence, then there exists an integer $c$ such that $$a_n = F_{n+k-2} + (-1)^nc.$$

We say that a prime $p$ is $\textit{philé}$ if there is a polynomial $P$ of non-negative integer coefficients smaller than $p$ and with degree $3$, that is, $P(x) = ax^3 + bx^2 + cx + d$ where $a, b, c, d < p$, such that $$\{P(n) | 1 \leq n \leq p\}$$is a complete residue system modulo $p$. Find all $\textit{philé}$ primes. Note: A set $A$ is a complete residue system modulo $p$ if for every integer $k$, with $0 \leq k \leq p - 1$, there exists an element $a \in A$ such that $$p | a-k.$$

Let $n \geq 2023$ be an integer. For each real $x$, we say that $\lfloor x \rceil$ is the closest integer to $x$, and if there are two closest integers then it is the greater of the two. Suppose there is a positive real $a$ such that $$\lfloor an \rceil = n + \bigg\lfloor\frac{n}{a} \bigg\rceil.$$Show that $|a^2 - a - 1| < \frac{n\varphi+1}{n^2}$.

We all know the Fibonacci sequence. However, a slightly less known sequence is the $k$-bonacci sequence. In it, we have $F_1^{(k)} = F_2^{(k)} = \cdots = F_{k-1}^{(k)} = 0, F_k^{(k)} = 1$ and $$F^{(k)}_{n+k} = F^{(k)}_{n+k-1} + F^{(k)}_{n+k-2} + \cdots + F^{(k)}_n,$$for all $n \geq 1$. Find all positive integers $k$ for which there exists a constant $N$ such that $$F^{(k)}_{n-1}F^{(k)}_{n+1} - (F ^{(k)}_n)^2 = (-1)^n$$for every positive integer $n \geq N$.