For all primes $p \geq 3,$ define $F(p) = \sum^{\frac{p-1}{2}}_{k=1}k^{120}$ and $f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\}$, where $\{x\} = x - [x],$ find the value of $f(p).$

Click for solution USe the lemma $1^m+2^m+..(p-1)^m \equiv 0 (mod p)$ for all $m$ satisfies $0 \leq m \leq (p-2)$

Let $n \geq 2, n \in \mathbb{N}$, $a,b,c,d \in \mathbb{N}$, $\frac{a}{b} + \frac{c}{d} < 1$ and $a + c \leq n,$ find the maximum value of $\frac{a}{b} + \frac{c}{d}$ for fixed $n.$

A graph $G=(V,E)$ is given. If at least $n$ colors are required to paints its vertices so that between any two same colored vertices no edge is connected, then call this graph ''$n-$colored''. Prove that for any $n \in \mathbb{N}$, there is a $n-$colored graph without triangles.

Click for solution From a k-chromatic triangle free graph G lets construct a k+1 chromatic triangle free graph G'. Let X(G) be the chromatic number of G. Let V(G)= {v1,...,vn}. Beginning with G add vertices X={x1,...,xn} and one more vertex y. Add edges to make x_i adjacent to all of N(v_i), and let N(y)= X. Clearly X is an idependet set in G'. Hence the other vertices of any triangle containing x_i belong to V(G) and are neighbors of v_i. This would complete a triangle in G, which can't exist. Therefore G' is triangle free. A proper k-coloring f of G extends to a proper k+1 coloring of G' by setting f(x_i)=f(v_i) and f(y)=k+1. Therefore X(G')<=X(G) + 1. So to obtain equality we show that X(G) < X(G'). So we consider any proper coloring of G' and obain from it a proper coloring of G using fewer colors. Let g be a proper k-coloring of G'. By changing the names of colors we may assume that g(y)=k. This restricts g to {1,...,k-1} on X. In V(G), it may use all k colors. Let A be the set of vertices in G on which g uses color k; we change the colors used on A to obain a proper k-1 coloring of G. For each v_1 in A we change the color of v_1 to g(x_i). Because all the vertices in A have color k under g, no two vertices of A are adjacent. Thus we need only check edges of the form v_i,v' with v_i in A and v' in V(G) - A. If v' is adjacent to v_i then by construction also v' is adjacent to x_i, which yields that g(v') and g(x_i) are different. Since we change the color on v_i to g(x_i) our change does not violate the edge v_i,v'. So the modified coloring of V(G) is a proper k-1 coloring of G.

Find all integer solutions to $2 x^4 + 1 = y^2.$

Click for solution For $|y|<2$ the solutions $(x,y)$ are $(0,-1),(0,1)$. Let's assume $y>1$, the equation is equivalent to $(y+1)(y-1)=2x^4$. Because $y+1,y-1$ are coprime for odd divisors, then there exist such odd, coprime numbers $u,v$ that $x=2^tuv$ and $y-1=2^mu^4$ , $y+1=2^nv^4$, where $m,n$ are postive integers and $t$ is a non-negative positive integer. Let's note that $m+n = 4t+1= 1$ $(\mod 4)$. So we have $2^nv^4-2^mu^4=2$, so $min\{m,n\}=1$ and $max\{m,n\}>min\{m,n\}$: 1) $m=1$ thus $2^{n-1}v^4-u^4=1$ ($n>1$), but because $m+n = 1$ $(\mod 4)$ so $n-1=3$ $(\mod 4)$,so $n \geq 3$. However, $u^4+1=2$ $(\mod 4)$, so we have a contradiction because $8|2^{n-1}$ so $2^{n-1}=0$ $(\mod 4)$. 2)$n=1$ thus $v^4-2^{m-1}u^4=1$ ($n>1$), so $(v^2-1)(v^2+1)=2^{m-1}u^4$. Because $v^2+1,v^2-1$ ($v>1$) are coprime for odd divisors, then there exist such odd, coprime numbers $a,b$ that $u=ab$ and $v^2-1=2^{m-2}a^4$ , $v^2+1=2b^4$. Thus $(v+2^{\frac{m-2}{2}}a^2)(v-2^{\frac{m-2}{2}}a^2)=1$ ( because $m+n=1$ $(\mod 4)$ and $n=1$ then $m-2$ is even), but that is impossible. 1) and 2) haven't given any solutions, so there aren't any solutions for $y>1$ and also for $y<-1$- the equality is symmetric, so the only solutions are: $(0,1),(0,-1)$.

Let $S = \{(x,y) | x = 1, 2, \ldots, 1993, y = 1, 2, 3, 4\}$. If $T \subset S$ and there aren't any squares in $T.$ Find the maximum possible value of $|T|.$ The squares in T use points in S as vertices.

Let $ABC$ be a triangle and its bisector at $A$ cuts its circumcircle at $D.$ Let $I$ be the incenter of triangle $ABC,$ $M$ be the midpoint of $BC,$ $P$ is the symmetric to $I$ with respect to $M$ (Assuming $P$ is in the circumcircle). Extend $DP$ until it cuts the circumcircle again at $N.$ Prove that among segments $AN, BN, CN$, there is a segment that is the sum of the other two.

Click for solution Noticing the 3 triangles are similar: AB'C' ~ A'BC'~ A'CB', we obtain The Euler line of of AB'C' intersects AB', AC', B'C' and forming respectively the acute angles $a, b, c$. The Euler line of of A'BC' intersects A'B, A'C', BC' and forming respectively the acute angles $a, b, c$. The Euler line of of A'CB' intersects A'C, A'B', B'C and forming respectively the acute angles $a, b, c$. We also have $O_aO_b // AB, O_cO_b // CB, O_aO_c // AC$. Hence applying the trig Ceva to triangle $O_aO_bO_c$, we 'll obtain our claim. Grobber 's problem is equivalent to mine by showing that A'B'C' have A, B, C its' excenters. Therefore, if you call $K_a, K_b, K_c$ such that there exists m, n, p such that $ mK_aA + nK_aB' + pK_aC' = 0 , mK_bA' + nK_bB + pK_bC' = 0, mK_cA' + nK_cC + pK_cB' = 0$ (they are all vectors) then $O_aK_a, O_bK_b, O_cK_c$ are concurent.