The 40 unit squares of the 9 9-table (see below) are labeled. The horizontal or vertical row of 9 unit squares is good if it has more labeled unit squares than unlabeled ones. How many good (horizontal and vertical) rows totally could have the table?

Let $ m,n$ be integers such that $ 0\le m\le 2n$. Then prove that the number $ 2^{2n + 2} + 2^{m + 2} + 1$ is perfect square iff $ m = n$.

Let $ A$ be a set of $ 2n$ points on the plane such that no three points are collinear. Prove that for any distinct two points $ a,b\in A$ there exists a line that partitions $ A$ into two subsets each containing $ n$ points and such that $ a,b$ lie on different sides of the line.

For the positive real numbers $ a,b,c$ prove the inequality \[ \frac {c}{a + 2b} + \frac {a}{b + 2c} + \frac {b}{c + 2a}\ge1. \]

Let the circle $ (I; r)$ be inscribed in the triangle $ ABC$. Let $ D$ be the point of contact of this circle with $ BC$. Let $ E$ and $ F$ be the midpoints of $ BC$ and $ AD$, respectively. Prove that the three points $ I$, $ E$, $ F$ are collinear.

Find all prime numbers $ p,q$ less than 2005 and such that $ q|p^2 + 4$, $ p|q^2 + 4$.

Click for solution If $ p$ or $ q$ is $ 2$, then we may let $ p = 2$, so $ q|8$, so $ q = 2$. This means that $ (p, q) = 2$. Otherwise, let $ p$ and $ q$ be odd primes. We have that $ q|(p^2 + 4)$, so $ p$ and $ q$ may not be equal. Therefore, $ q|(p^2 + 4)$ and $ p|(q^2 + 4)$ if and only if $ \frac {p^2q^2 + 4p^2 + 4q^2 + 16}{pq}$ is an integer. This is an integer if and only if $ \frac {4p^2 + 4q^2 + 16}{pq}$ is an integer. Since $ p$ and $ q$ are relatively prime with $ 4$, we see that this is an integer if and only if $ \frac {p^2 + q^2 + 4}{pq}$ is an integer. Let this value be $ k$. Then, we have that $ p^2 - p(kq) + (q^2 + 4) = 0$. In general, let us consider the equation $ a^2 - a(bk) + (b^2 + 4) = 0$. Let $ (a, b)$ be the the solution to this equation in positive integers with the lowest possible $ a + b$, and without loss of generality, say that $ a\le b$. Now, if $ (a, b)$ is a solution, then $ (\frac {b^2 + 4}{a}, b)$ is a solution, so $ (b, \frac {b^2 + 4}{a})$ must be a solution. Thus, $ b + \frac {b^2 + 4}{a} \ge a + b$, so $ a^2 - b^2 \le 4\implies (a + b)(a - b) \le 4$. Now, this means that $ a = b$ or $ (a, b)$ is $ (1, 2)$. If $ a = b$, then we have that $ (k - 2)a^2 = 4$, so $ (a, b)$ is $ (1,1)$ or $ (2,2)$. We find $ k$ for each of these cases. If $ (a, b) = (1, 2)$, then $ k$ is not an integer. If $ (a, b) = (1,1)$, then $ k = 6$. If $ (a, b) = (2, 2)$, then $ k = 3$. If $ k = 3$, then we have that $ \frac {p^2 + q^2 + 4}{pq}$ is odd, which is impossible since the numerator is even and the denominator is odd. It follows that $ k = 6$. Thus, $ p^2 - 6pq + (q^2 + 4) = 0$. The discriminant of this is $ 16(2q^2 - 1)$, so $ 2q^2 - 1$ is a perfect square. We now consider the Pell's Equation $ 2m^2 - n^2 = 1$. We can compute the primes $ q$'s that work for the the Pell's Equation under $ 2005$ (they are solutions to the recurrence $ a_{n + 1} = 6a_n - a_{n - 1}$ with $ a_1 = 1$ and $ a_2 = 5$), which are $ 5$ and $ 29$. Plugging this in, we get that $ p^2 - 30p + 29 = 0$, so $ p = 29$ or $ p = 1$ (but $ p = 1$ is impossible since $ 1$ is not prime, so $ p = 29$). If $ q = 29$, then $ p^2 - 174p + 845 = 0$, so $ (p - 169)(p - 5) = 0$, so $ p = 5$ or $ p = 169$ (but $ p = 169$ is impossible as $ 169$ is composite). Thus, the only solutions are $ (p, q) = (5, 29), (29, 5), (2,2)$.

Prove that the equation $ x^{5} + 31 = y^{2}$ has no integer solution.

Let $ r$ be a real number such that the sequence $ (a_{n})_{n\geq 1}$ of positive real numbers satisfies the equation $ a_{1} + a_{2} + \cdots + a_{m + 1} \leq r a_{m}$ for each positive integer $ m$. Prove that $ r \geq 4$.

Let SABC be a regular triangular pyramid. Find the set of all points $ D (D! = S)$ in the space satisfing the equation $ |cos ASD - 2cosBSD - 2 cos CSD| = 3$.

For the positive real numbers $ a,b,c$ prove that \[ \frac c{a + 2b} + \frac d{b + 2c} + \frac a{c + 2d} + \frac b{d + 2a} \geq \frac 43.\]

The inner point $ X$ of a quadrilateral is observable from the side $ YZ$ if the perpendicular to the line $ YZ$ meet it in the colosed interval $ [YZ].$ The inner point of a quadrilateral is a $ k-$point if it is observable from the exactly $ k$ sides of the quadrilateral. Prove that if a convex quadrilateral has a 1-point then it has a $ k-$point for each $ k=2,3,4.$

Find all prime numbers $ p,q < 2005$ such that $ q | p^{2} + 8$ and $ p|q^{2} + 8.$

Click for solution Solution: Clearly $ p=q=2$ satisfies to the conditions.Now assume that $ p,q\geq 3$. As we know $ \frac{(p^2+8)(q^2+8)}{pq}$ is an integer. It follows that $ \frac{8(p^2+q^2+8)}{pq}$, recalling that $ p,q$ are odd, it can be inferred that $ \frac{p^2+q^2+8}{pq}$ is an integer as well. So the problem reduces to finding all odd solutions $ (p_n,q_n)$, that are less than $ 2005$ and satisfy to condition, that: \[ \frac{p^2+q^2+8}{pq}=k \] for some fixed $ k$. Assume that $ (p_0,q_0)$ is the solution, so that the sum $ p+q$ reaches its minimum, and $ p_0\geq q_0$. If $ p_0=q_0$, then, obviously $ p_0=q_0=1$. Now assume $ p_0>q_0$. It follows that $ p_0^2-p_0(q_0k)+q_0^2+8=0$ As we know, by Vieta's formula: $ p'=\frac{q_0^2+8}{p_0}<\frac{p_0^2-2p_0+9}{p_0}<p_0$, as long as, $ p_0\geq 5$, so this case leads to contradiction. So we might assume that $ p_0\leq 4$. If $ p_0=3$, then $ q_0=1$, but this case is invalid. So if $ (p_0,q_0)$ is a solution than $ p_0=q_0=1$ and $ k=10$. All solutions determined by this method are in the sequence $ q_{n+1}=p_n, p_{n+1}=10q_n-p_n$ are $ (1,1),(9,1),(89,9),(881,89)$. Now let's prove that all solutions of the equation $ p^2-10pq+q^2+8=0$ are in the described sequence. Let $ (p',q')$ be the odd solution with minimal sum that is not in the sequence. By the explicitly described method, we could proceed to the another solution with less sum and that will either lead to the contradiction of minimality or to the contradiction of the assumptation that this solution is not in the sequence. So all solutions are: \[ (2,2),(881,89),(89,881) \]