Given a graph with $n$ ($n\ge 4$) vertices . It is known that for any two vertices $A$ and $B$ there exists a vertex which is connected by edges both with $A$ and $B$. Find the smallest possible numbers of edges in the graph. E. Barabanov

Fibonacci numbers are defined as follows: $F_0 = F_1 = 1, F_{n+2} = F_{n+1}+F_n, n \geq 0$. Let $a_n$ be the number of words that consist of $n$ letters $0$ or $1$ and contain no two letters $1$ at distance two from each other. Express $a_n$ in terms of Fibonacci numbers.

Rectangle $ABCD$ has $AB = 24$ and $BC = 7$. Let $d$ be the distance between the centers of the incircles of $\vartriangle ABC$ and $\vartriangle CDA$. Find $d^2$.

Let $k$ be a positive integer such that $p = 8k + 5$ is a prime number. The integers $r_1, r_2, \dots, r_{2k+1}$ are chosen so that the numbers $0, r_1^4, r_2^4, \dots, r_{2k+1}^4$ give pairwise different remainders modulo $p$. Prove that the product \[\prod_{1 \leqslant i < j \leqslant 2k+1} \big(r_i^4 + r_j^4\big)\] is congruent to $(-1)^{k(k+1)/2}$ modulo $p$. (Two integers are congruent modulo $p$ if $p$ divides their difference.) [i]Fedor Petrov[/i]

Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\angle ABC=\angle DEA=90^\circ$ and $BC+DE=1$. Compute the area of the pentagon. [i]Greece[/i]

Quadratic equation $ x^2\plus{}ax\plus{}b\plus{}1\equal{}0$ have 2 positive integer roots, for integers $ a,b$. Show that $ a^2\plus{}b^2$ is not a prime.

In how many ways can the set $N ={1, 2, \cdots, n}$ be partitioned in the form $p(N) = A_{1}\cup A_{2}\cup \cdots \cup A_{k},$ where $A_{i}$ consists of elements forming arithmetic progressions, all with the same common positive difference $d_{p}$ and of length at least one? At least two? [hide="Solution"] [b]Part 1[/b] Claim: There are $2^{n}-2$ ways of performing the desired partitioning. Let $d(k)$ equal the number of ways $N$ can be partitioned as above with common difference $k.$ We are thus trying to evaluate $\sum_{i=1}^{n-1}d(i)$ [b]Lemma: $d(i) = 2^{n-i}$[/b] We may divide $N$ into various rows where the first term of each row denotes a residue $\bmod{i}.$ The only exception is the last row, as no row starts with $0$; the last row will start with $i.$ We display the rows as such: $1, 1+i, 1+2i, 1+3i, \cdots$ $2, 2+i, 2+2i, 2+3i, \cdots$ $\cdots$ $i, 2i, 3i, \cdots$ Since all numbers have one lowest remainder $\bmod{i}$ and we have covered all possible remainders, all elements of $N$ occur exactly once in these rows. Now, we can take $k$ line segments and partition a given row above; all entries within two segments would belong to the same set. For example, we can have: $1| 1+i, 1+2i, 1+3i | 1+4i | 1+5i, 1+6i, 1+7i, 1+8i,$ which would result in the various subsets: ${1},{1+i, 1+2i, 1+3i},{1+4i},{1+5i, 1+6i, 1+7i, 1+8i}.$ For any given row with $k$ elements, we can have at most $k-1$ segments. Thus, we can arrange any number of segments where the number lies between $0$ and $k-1$, inclusive, in: $\binom{k-1}{0}+\binom{k-1}{1}+\cdots+\binom{k-1}{k-1}= 2^{k-1}$ ways. Applying the same principle to the other rows and multiplying, we see that the result always gives us $2^{n-i},$ as desired. We now proceed to the original proof. Since $d(i) = 2^{n-i}$ by the above lemma, we have: $\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}= 2^{n}-2$ Thus, there are $2^{n}-2$ ways of partitioning the set as desired. [b]Part 2[/b] Everything is the same as above, except the lemma slightly changes to $d(i) = 2^{n-i}-i.$ Evaluating the earlier sum gives us: $\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$ [/hide]

[b]a)[/b] Let $ a,b,c $ be three complex numbers. Prove that the element $ \begin{pmatrix} a & a-b & a-b \\ 0 & b & b-c \\ 0 & 0 & c \end{pmatrix} $ has finite order in the multiplicative group of $ 3\times 3 $ complex matrices if and only if $ a,b,c $ have finite orders in the multiplicative group of complex numbers. [b]b)[/b] Prove that a $ 3\times 3 $ real matrix $ M $ has positive determinant if there exists a real number $ \lambda\in\left( 0,\sqrt[3]{4} \right) $ such that $ A^3=\lambda A+I. $ [i]Cristinel Mortici[/i]

A certain frog that was placed on a vertex of a convex polygon chose to jump to another vertex, either clockwise skipping one vertex, either counterclockwise skipping two vertexes, and repeated the procedure. If the number of jumps that the frog made is equal to the number of sides of the polygon, the frog has passed through all its vertexes and ended up on the initial vertex, what´s the set formed by all the possible values that this number can take? [i]Andrei Eckstein[/i]

Let $\omega=\cos\frac{2\pi}{7}+i\cdot\sin\frac{2\pi}{7}$, where $i=\sqrt{-1}$. Find $$\prod_{k=0}^{6}(\omega^{3k}+\omega^k+1).$$

There are $11$ empty boxes. In one move you can put one coin in some 10 of them. Two people play and take turns. Wins the one after which for the first time there will be $21$ coins in one of the boxes. Who wins when played correctly?

If $p$ is a prime positive integer and $x,y$ are positive integers, find , in terms of $p$, all pairs $(x,y)$ that are solutions of the equation: $p(x-2)=x(y-1)$. (1) If it is also given that $x+y=21$, find all triplets $(x,y,p)$ that are solutions to equation (1).

Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xf(x-y))+yf(x)=x+y+f(x^2),$$ for all real numbers $x$ and $y.$

We call a tetrahedron right-faced if each of its faces is a right-angled triangle. [i](a)[/i] Prove that every orthogonal parallelepiped can be partitioned into six right-faced tetrahedra. [i](b)[/i] Prove that a tetrahedron with vertices $A_1,A_2,A_3,A_4$ is right-faced if and only if there exist four distinct real numbers $c_1, c_2, c_3$, and $c_4$ such that the edges $A_jA_k$ have lengths $A_jA_k=\sqrt{|c_j-c_k|}$ for $1\leq j < k \leq 4.$

Show that $\sigma (n) -d(m)$ is even for all positive integers $m$ and $n$ where $m$ is the largest odd divisor of $n$.

Octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle where $A_1A_2=A_2A_3=A_3A_4=A_6A_7=13$ and $A_4A_5=A_5A_6=A_7A_8=A_8A_1=5. $ The sum of all possible areas of $A_1A_2A_3A_4A_5A_6A_7A_8$ can be expressed as $a+b\sqrt{c}$ where $\gcd{a,b}=1$ and $c$ is squarefree. Find $abc.$ [asy] label("$A_1$",(5,0),E); label("$A_2$",(2.92, -4.05),SE); label("$A_3$",(-2.92,-4.05),SW); label("$A_4$",(-5,0),W); label("$A_5$",(-4.5,2.179),NW); label("$A_6$",(-3,4), NW); label("$A_7$",(3,4), NE); label("$A_8$",(4.5,2.179),NE); draw((5,0)--(2.9289,-4.05235)); draw((2.92898,-4.05325)--(-2.92,-4.05)); draw((-2.92,-4.05)--(-5,0)); draw((-5,0)--(-4.5, 2.179)); draw((-4.5, 2.179)--(-3,4)); draw((-3,4)--(3,4)); draw((3,4)--(4.5,2.179)); draw((4.5,2.179)--(5,0)); dot((0,0)); draw(circle((0,0),5)); [/asy]

A beam of light shines from point $L$, reflects off a reflector at point $S$, and reaches point $D$ so that $\overline{SD}$ is perpendicular to $\overline{ML}$. Then $x$ is [DIAGRAM NEEDED] $ \mathrm{(A) \ } 13^\circ \qquad \mathrm{(B) \ } 26^\circ \qquad \mathrm {(C) \ } 32^\circ \qquad \mathrm{(D) \ } 58^\circ \qquad \mathrm{(E) \ } 64^\circ$

Let $a_1, a_2... $ be an infinite sequence of real numbers such that $a_{n+1}=\sqrt{{a_n}^2+a_n-1}$. Prove that $a_1 \notin (-2,1)$ [i]Proposed by Oleg Mushkarov and Nikolai Nikolov [/i]

On each non-boundary unit segment of an $8\times 8$ chessboard, we write the number of dissections of the board into dominoes in which this segment lies on the border of a domino. What is the last digit of the sum of all the written numbers?

Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.

Find the maximum value of $k$ for which one can choose $k$ integers out of $1,2... ,2n$ so that none of them divides another one.

A [i]lazy[/i] rook can only move from a square to a vertical or a horizontal neighbour. It follows a path which visits each square of an $8 \times 8$ chessboard exactly once. Prove that the number of such paths starting at a corner square is greater than the number of such paths starting at a diagonal neighbour of a corner square. [i](7 points)[/i]

Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $D$ be the midpoint of the minor arc $AB$ of $\Omega$. A circle $\omega$ centered at $D$ is tangent to $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $ \Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$. Poland

2017 prime numbers $p_1,...,p_{2017}$ are given. Prove that $\prod_{i<j} (p_i^{p_j}-p_j^{p_i})$ is divisible by 5777.

Let \(m\) be a positive integer. Find, in terms of \(m\), all polynomials \(P(x)\) with integer coefficients such that for every integer \(n\), there exists an integer \(k\) such that \(P(k)=n^m\). [i]Proposed by Raymond Feng[/i]