be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
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 
,
,
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 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
intersects line 
.
.
can be placed tangent to side 
of 
,
.
circles of radius 
can also be placed along 
,
and 
.
,
.
subsets of 
is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set 
,
for the number of subsets of 
that are blue.
and 
of ![\[ f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2). \]](http://latex.artofproblemsolving.com/1/8/1/18184e9196e6c73601653d85f1cfd6d9798a8e43.png)
Y by buratinogigle, adityaguharoy, Adventure10, Mango247
Let 
be a triangle with circumcenter 
and circumcircle 
The point 
lies on 
such that 
is the 
- symmedian of triangle 
The line through perpendicular to intersects 
in 
respectively. Denote by 
the nine-point circle of triangle 
and let and intersect again in 
Further, let the tangent to at meet the line 
in 
and let 
be the antipode of with respect to circle Prove that the points 
are collinear.
Notes: 1. The-symmedian of triangle is the reflection of the -median in the -angle bisector.
2. The antipode of a point with respect to a circle is the point on the circle diametrically opposite to it.
Proposed by Adithya Bhaskar
be a triangle with circumcenter 
and circumcircle 
The point 
lies on 
such that 
is the 
- symmedian of triangle 
The line through perpendicular to intersects 
in 
respectively. Denote by 
the nine-point circle of triangle 
and let and intersect again in 
Further, let the tangent to at meet the line 
in 
and let 
be the antipode of with respect to circle Prove that the points 
are collinear.Notes: 1. The
-symmedian of triangle is the reflection of the -median in the -angle bisector.2. The antipode of a point with respect to a circle is the point on the circle diametrically opposite to it.
Proposed by Adithya Bhaskar
This post has been edited 2 times. Last edited by AdithyaBhaskar, May 7, 2016, 4:55 PM
is the 
and since 
then 
at 
Therefore, 
at their midpoints 
is 9-point circle of 
be the 2nd intersection of 
with 
is tangent to 
Thus if 
we have 
is midpoint of 
Together with 
we have 
are collinear.
inversion reduces this to an intermixing of Russian MO 2015 10th grade P7 and IMO Shortlist 2011 G4.
be the midpoints of 
is harmonic, 
is tangent to 
,
,
.
and swaps 
,
.
be the inverses of 
.
is harmonic, 
is the midpoint of 
.
.
lies on the perpendicular bisector of 
,
.
