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.
be an integer. We consider a square grid of size 
divided into 
unit squares. The grid is called balanced if:
, 
or 
.
.
, the smallest integer 
such that any balanced grid always contains an 
square whose absolute sum of the 
cells is less than or equal to 
.
of distinct positive real numbers is called radiant if it satisfies the following property: if 
and 
are two distinct elements of , then 
is also an element of .
?
or 
.
.
be an acute-angled triangle in which 
and 
. Let point 
lie on segment 
and point 
lie on segment 
such that 
, 
and 
. Let 
be the circumcenter of triangle 
, 
the orthocenter of triangle , and 
the point of intersection of the lines 
and 
. Prove that , , and are collinear.
consecutive even numbers, such that the square of their product divides the sum of their squares.
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
an increasing sequence, with 
, if 
then 
increasing, from a certain rank.
?
the points 
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
such that 
. Show that 
. When will equality hold?
is not a field, the problem is faulty
is a finite field for all primes 
.
and 
.
. The geometric multiplicity is 1.
and 
. If we take instead 1 of column 4 and 3 of column 5, which means we have to take none of column 2. So another vector in this kernel is 
.
. We can take instead 1 of column 3 and 2 of column 4. In that case, we can take 1 of column 1 and 3 of column 2. This gives us 
.
, 
, 
, 
, and 
where:
, 
, 
over the field 
:
instead of 