Prove whether or not there exist natural numbers $n,k$ where $1\le k\le n-2$ such that \[\binom{n}{k}^2+\binom{n}{k+1}^2=\binom{n}{k+2}^4 \]
2011 Bulgaria National Olympiad
Day 1
Let $f_1(x)$ be a polynomial of degree $2$ with the leading coefficient positive and $f_{n+1}(x) =f_1(f_n(x))$ for $n\ge 1.$ Prove that if the equation $f_2(x)=0$ has four different non-positive real roots, then for arbitrary $n$ then $f_n(x)$ has $2^n$ different real roots.
Triangle $ABC$ and a function $f:\mathbb{R}^+\to\mathbb{R}$ have the following property: for every line segment $DE$ from the interior of the triangle with midpoint $M$, the inequality $f(d(D))+f(d(E))\le 2f(d(M))$, where $d(X)$ is the distance from point $X$ to the nearest side of the triangle ($X$ is in the interior of $\triangle ABC$). Prove that for each line segment $PQ$ and each point interior point $N$ the inequality $|QN|f(d(P))+|PN|f(d(Q))\le |PQ|f(d(N))$ holds.
Day 2
Point $O$ is inside $\triangle ABC$. The feet of perpendicular from $O$ to $BC,CA,AB$ are $D,E,F$. Perpendiculars from $A$ and $B$ respectively to $EF$ and $FD$ meet at $P$. Let $H$ be the foot of perpendicular from $P$ to $AB$. Prove that $D,E,F,H$ are concyclic.
For each natural number $a$ we denote $\tau (a)$ and $\phi (a)$ the number of natural numbers dividing $a$ and the number of natural numbers less than $a$ that are relatively prime to $a$. Find all natural numbers $n$ for which $n$ has exactly two different prime divisors and $n$ satisfies $\tau (\phi (n))=\phi (\tau (n))$.
In the interior of the convex 2011-gon are $2011$ points, such that no three among the given $4022$ points (the interior points and the vertices) are collinear. The points are coloured one of two different colours and a colouring is called "good" if some of the points can be joined in such a way that the following conditions are satisfied: 1) Each segment joins two points of the same colour. 2) None of the line segments intersect. 3) For any two points of the same colour there exists a path of segments connecting them. Find the number of "good" colourings.