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.
, let 
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer such that 
.
let 
be the foot of the altitude from 
to the side 
and 
, 
, 
be the incenter, -excenter, and 
-excenter, respectively. Denote by 
and 
the other intersection points of the circle 
with the lines 
and 
, respectively. Prove that 
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
, such that 
and 
.
such that 
is prime, 
with 
and ?
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
be a triangle, and let 
and 
be points on segment symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line at . Similarly, let 
denote the circle passing through and tangent to line at . Let the circles and intersect again at point 
(
). Prove that 
.
of 
cuts line 
again at a point 
Since 
line bisects both the angles 
and 
This is possible only when 
which means that 
are A-isosceles, or when points 
are identical, which means that quadrilateral 
is cyclic. The first possibility is eliminated by the condition that 
is scalene. of 
cuts 
at 
and again at a point 
Denote 
On one hand, powers of 
to are
and lower signs for 
.
and the two expressions for 
.resp. 
is obtuse.
and 
be points on 
such that 
are angle bisectors of the triangle, respectively. Assume that 
Prove that
and 
on 
,