AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
, 
, 
, 
such that 
and 
for 
.
, which is a perfect square of an integer?
, prove that there exists a positive integer 
, such that for any integer 
, can be expressed by a sum of positive integers 
's as![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
, 
, 
, 
.
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be a function such that
for all 
. Determine all such functions 
.
such that the equality 
holds for all pairs of real numbers 
.
with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
. and an integer , 
is defined as follows: 
if 
and 
if 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
be a positive integer. A positive integer is called a benefactor of if the positive divisors of can be partitioned into two sets 
and 
such that is equal to the sum of elements in minus the sum of the elements in . Note that or could be empty, and that the sum of the elements of the empty set is 
.
is a benefactor of 
because 
. has at least 
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained?
be a cyclic convex quadrilateral, where 
is an equilateral triangle. Denote 
and 
Prove that ![$[KLM]=2\cdot [ABC],$](http://latex.artofproblemsolving.com/f/b/0/fb00b2478efb3c51bcbd648643288cf6b1c7b746.png)
where denoted by 
- the area of 
is the Fermat Point of 
Hence ![$[KLM]=[KDL]+[LDM]+[MDK]=\dfrac{\sqrt{3}}{4}\left(DK\cdot DL+DL\cdot DM+DM\cdot DK\right).$](http://latex.artofproblemsolving.com/e/3/c/e3c18e6377c41bfcfa05ab589df10d73996a2889.png)
we have 
and similarly 
so it suffices to show that 
so 
by Law of Cosines in 
, we are motivated to length bash. WLOG 
. Let 
. Let 
.
. 
is the angle bisector of 
.![$[ABC]=\frac{AB\cdot BC}{BL \cdot BM}[BLM]$](http://latex.artofproblemsolving.com/8/d/1/8d1c19716b017386b0fa081cddd0cac9da940fcf.png)
and ![$[KLM]=\frac{KP}{BP}[BLM]$](http://latex.artofproblemsolving.com/e/3/b/e3b9fbe72df14c2c3aef2a4a31e70ff1bb5a5760.png)
, similarly, 
.
.
.
![$\frac{[KML]}{[ABC]}=\frac{\frac{PK}{BP}[BLM]}{\frac{AB\cdot BC}{BL \cdot BM}[BLM]}=\frac{2x(1-x)}{x(1-x)}=\boxed{2} \text{Q.E.D.} \Box$](http://latex.artofproblemsolving.com/8/b/1/8b172f0a65c8e3212fbbc2608af521ae0e51e0a1.png)
denote the complex conjugate of 
Let 
lie on the unit circle where 
is a primitive third root of unity. As before, ![$[KLM]=\dfrac{\sqrt{3}}{4}(DK\cdot DL+DL\cdot DM+DM\cdot DK).$](http://latex.artofproblemsolving.com/d/f/d/dfdffee32eff7ffa08505c42b472e7c09253c566.png)
and 
of the unit circle are given by 
we easily obtain 
and similarly 
so 
as desired.
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
