Find all pairs of positive integers $(m,n)$ for which there exist relatively prime integers $a$ and $b$ greater than $1$ such that $$\frac{a^m+b^m}{a^n+b^n}$$ is an integer.

Prove that there exists an infinite sequence of positive integers $a_1,a_2,a_3,\dots$ such that for any positive integer $k$, $a_k^2+a_k+2023$ has at least $k$ distinct positive divisors.

A polynomial $ P(x)$ with integer coefficients is called good,if it can be represented as a sum of cubes of several polynomials (in variable $ x$) with integer coefficients.For example,the polynomials $ x^3 \minus{} 1$ and $ 9x^3 \minus{} 3x^2 \plus{} 3x \plus{} 7 \equal{} (x \minus{} 1)^3 \plus{} (2x)^3 \plus{} 2^3$ are good. a)Is the polynomial $ P(x) \equal{} 3x \plus{} 3x^7$ good? b)Is the polynomial $ P(x) \equal{} 3x \plus{} 3x^7 \plus{} 3x^{2008}$ good? Justify your answers.

Find all triples of positive integers $(a,b,c)$ such that: $a^2bc-2ab^2c-2abc^2+b^3c+bc^3+2b^2c^2=11$

For each positive integer $n$, let $S_n = \sum_{k=1}^nk^3$, and let $d(n)$ be the number of positive divisors of $n$. For how many positive integers $m$, where $m\leq 25$, is there a solution $n$ to the equation $d(S_n) = m$?

Find all complex numbers $m$ such that polynomial \[x^3 + y^3 + z^3 + mxyz\] can be represented as the product of three linear trinomials.

Let $n$ be a positive integer. If $4^n + 2^n + 1$ is a prime, prove that $n$ is a power of three.

Let be a natural number $ n $ and $ n $ real numbers $ a_1,a_2,\ldots ,a_n $ such that $$ a_m+a_{m+1} +\cdots +a_n\ge \frac{(m+n)(n-m+1)}{2} ,\quad\forall m\in\{ 1,2,\ldots ,n \} . $$ Prove that $ a_1^2+a_2^2+\cdots +a_n^2\ge\frac{n(n+1)(2n+1)}{6} . $

Let $n$ be a positive integer. If $4^n + 2^n + 1$ is a prime, prove that $n$ is a power of three.

Given $a, \theta \in \mathbb R, m \in \mathbb N$, and $P(x) = x^{2m}- 2|a|^mx^m \cos \theta +a^{2m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.

Three integers are given. $A$ denotes the sum of the integers, $B$ denotes the sum of the square of the integers and $C$ denotes the sum of cubes of the integers(that is, if the three integers are $x, y, z$, then $A=x+y+z$, $B=x^2+y^2+z^2$, $C=x^3+y^3+z^3$). If $9A \geq B+60$ and $C \geq 360$, find $A, B, C$.

Decompose the expression into real factors: \[E = 1 - \sin^5(x) - \cos^5(x).\]

Let $a_1, a_2, \ldots, a_n$ be $n$ positive integers, and let $b_1, b_2, \ldots, b_m$ be $m$ positive integers such that $a_1 a_2 \cdots a_n = b_1 b_2 \cdots b_m$. Prove that a rectangular table with $n$ rows and $m$ columns can be filled with positive integer entries in such a way that * the product of the entries in the $i$-th row is $a_i$ (for each $i \in \left\{1,2,\ldots,n\right\}$); * the product of the entries in the $j$-th row is $b_j$ (for each $i \in \left\{1,2,\ldots,m\right\}$).

Let $x$ be a complex number such that $x^{2011}=1$ and $x\neq 1$. Compute the sum \[\dfrac{x^2}{x-1}+\dfrac{x^4}{x^2-1}+\dfrac{x^6}{x^3-1}+\cdots+\dfrac{x^{4020}}{x^{2010}-1}.\]

The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.

If real numbers $a>b>1$ satisfy the inequality$$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$$what is the minimum possible value of $\dfrac{\sqrt{a-b}}{b-1}$

Determine all polynomials $P(x)$ with integer coefficients such that $P(n)$ is divisible by $\sigma(n)$ for all positive integers $n$. (As usual, $\sigma(n)$ denotes the sum of all positive divisors of $n$.)

After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists. Joshua walks directly over the mural without paying any attention. "You're a square, Josh." "No, $\textit{you're}$ a square," retorts Joshua. "In fact, you're a $\textit{cube}$, which is $50\%$ freakier than a square by dimension. And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being." "Okay, you're a cubist caricature of male immaturity," asserts Alexis. Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella. He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*} Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers. For instance, \begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*} Josh begins to wonder if there is a smallest value of $n$ such that \[(a+b+c)^3\leq n(a^3+b^3+c^3)\] for all natural numbers $a$, $b$, and $c$. Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof. What is the correct value of $n$ that Joshua found?

Factor completely the expression $(a-b)^3+(b-c)^3+(c-a)^3$

Alice is once again very bored in class. On a whim, she chooses three primes $p$, $q$, $r$ independently and uniformly at random from the set of primes at most 30. She then calculates the roots of $px^2+qx+r$. What is the probability that at least one of her roots is an integer?

Find all the triplets of real numbers $(x , y , z)$ such that : $y=\frac{x^3+12x}{3x^2+4}$ , $z=\frac{y^3+12y}{3y^2+4}$ , $x=\frac{z^3+12z}{3z^2+4}$

Find all non-negative integers $x, y$ such that\[x^3y+x+y=xy+2xy^2\]

Let $a_1, a_2, \ldots, a_n$ be $n$ positive integers, and let $b_1, b_2, \ldots, b_m$ be $m$ positive integers such that $a_1 a_2 \cdots a_n = b_1 b_2 \cdots b_m$. Prove that a rectangular table with $n$ rows and $m$ columns can be filled with positive integer entries in such a way that * the product of the entries in the $i$-th row is $a_i$ (for each $i \in \left\{1,2,\ldots,n\right\}$); * the product of the entries in the $j$-th row is $b_j$ (for each $i \in \left\{1,2,\ldots,m\right\}$).

Prove that there are infinitely many positive integers $m$, such that $n^4+m$ is not prime for any positive integer $n$.