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.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
of injective functions from the set of real numbers to itself satisfying
and 
. (We say a function 
is injective if 
for any distinct real numbers 
and 
.)
, find the smallest integer 
such that any graph with vertices and at least edges must contain at least two disjoint cycles (cycles with no common vertices).
{
} consisting 
distinct positive integers is 
if for every 
the number 
is divisible by the product of all numbers in except 
. Find all integers such that exists a set consisting of 
positive integers.
with 
and 
.
?
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
be the Fermat point of a 
. Prove that the Euler line of the triangles 
, 
, 
are concurrent and the point of concurrence is 
, the centroid of .
, such that for 
? be a cyclic quadrilateral for which 
. Construct two squares 
and 
externally on the sides 
and 
respectively.
meets 
at 
, 
meets 
at 
and 
denotes circumcenter of . Prove that 
and 
meets again on 
.
is defined by 
and, for each 
, the number 
is the smallest positive integer greater than 
that has the following property: for each 
, the number 
is not a perfect square. Prove that, for all , it holds that 
.
. Now suppose that 
for 
where 
. Let's prove that 
. Consider 
and 
. 
and have no common element. Assume that 
is not in . 
and 
hence there must exists some such that both 
and 
are perfect squares for 
.![\[\sqrt{(N-2)^2+l+m+a_i}\geq \sqrt{(N-2)^2+l+a_i}+1\]](http://latex.artofproblemsolving.com/3/c/0/3c082456e591ba185d3e199ecc3f963fa3dbfc10.png)
![\[(N-2)^2+l+m+a_i\geq (N-2)^2+l+a_i+1+2\sqrt{(N-2)^2+l+a_i}\]](http://latex.artofproblemsolving.com/9/3/b/93b4df6d3e72787bdfab8d1efa423db8a5f5a348.png)
![\[2N-4>m\geq 1+2\sqrt{(N-2)^2+l+a_i}\geq 1+2(N-1)\geq 2N-1\]](http://latex.artofproblemsolving.com/5/d/a/5da118493cf2a5703b1b6d374b0d9bd9f6930052.png)
Which is impossible as desired.
. It comes down to a single lemma:
for wich 
, for each 
.
we would have: 
.
, but that's a contradiction because 
. 
such 
(
to be precise) there must be at least one 
for which there is no with 
and that means , as we wanted.
![\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]](http://latex.artofproblemsolving.com/9/5/9/9591f7c842d1b6afcba11b15b3ea3f9e01bc937a.png)
![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
