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.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
+1 w
be real numbers such that 
find the largest value of 
such that 
and 
are all perfect square.
be pairwise coprime positive integers with 
being prime and 
. On the segment ![$I = [0, a_1 a_2 \cdots a_n ]$](http://latex.artofproblemsolving.com/a/2/b/a2bce96b048c9b8fc94926db80aba37fd5037b4a.png)
of the real line, mark all integers that are divisible by at least one of the numbers 
. These points split 
into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by .
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
be a regular polygon, and let 
be its set of vertices. Each point in is colored red, white, or blue. A subset of is patriotic if it contains an equal number of points of each color, and a side of is dazzling if its endpoints are of different colors. is patriotic and the number of dazzling edges of is even. Prove that there exists a line, not passing through any point in , dividing into two nonempty patriotic subsets.
of natural numbers such that
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
. Suppose that:
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines and respectively.
and 
intersect on the line .
be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be on the minor arc . Let 
be tangents to the incircle, with 
on the circumcircle and 
is on the arc 
and 
is on the arc 
. Then it suffices to show 
is tangent to the incircle. By Pascal's on hexagon 
, we get the intersection of 
, call it 
, the intersection of 
, call it 
and the intersection of 
, call it 
, are collinear. In short, 
are collinear. Now use the converse of Brianchon on 
, since 
are concurrent, and 
are tangent to the incircle, we have is tangent to the incircle, and we are done.
, 
be the circumcircle and inscribed circles of triangl
,we can construct a triangle 
such that 
