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.
with 
and 
.
?
of positive integers such that
be a circle inscribed inside a rhombus 
. Let 
and 
be variable points on 
and 
respectively, such as 
is always the tangent line to . and the area of triangle 
is the same.
, 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).
passing through point 
is an oblique strophoid. 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 
.
be an acute triangle and let 
be the foot of the -bisector. Moreover, let 
be the midpoint of . The circle 
with diameter 
meets 
at 
, while the circle 
with diameter meets 
at 
. Assume that and lie inside . Prove that 
, , , 
are concyclic.
of degree 
having 
distinct real roots (
), such that for every root 
: 
.
defined on all real numbers and taking real values such that ![\[f(f(y)) + f(x - y) = f(xf(y) - x),\]](http://latex.artofproblemsolving.com/f/5/2/f528fc4047e21247663d4993913eadf346315be7.png)
for all real numbers 
. Prove that there does not exist a pair of rational numbers 
such that 
.
![$P\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/4/c/74cb2247069abb8a22666e87fcbd69a55a015004.png)
for which there exists 
such that for all 
we have: ![\[P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).\]](http://latex.artofproblemsolving.com/9/0/9/90954cdeade9d13149e1aaa03df2c87dff157041.png)
so that has the same number of digits as its number of different prime factors and the sum of these different prime factors is equal to the sum of exponents of all these primes in factorization of .
so that has the same number of digits as its number of different prime factors and the sum of these different prime factors is equal to the sum of exponents of all these primes in factorization of .
be the number of digits of .
( the prime factorization of )
.
.
.
and then 
.
.
is the only solution in this case.
.
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
