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)
, 
such that 
is a perfect square.
, such that is a perfect square.
which can never be perfect square.
Hence 
is only solution.
and 
for 
, the expression for will be congruent to 
, which can never be a perfect square. This puts a very nice bound for 
, and it's easy to check the case of 
.
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
is a natural number, here is my solution:
we have 
so this number is not a perfect square.
If 
we have to find 
is a perfect square. By taking modulo 
we see that 
and 
)
and this implies 
and 
We can only have 
and 
so in this case 
we have to find 
is a perfect square. Let 
This leads to 
Since 
it follows that we can only have 
and 
so 
and 
so in this case 
If 
,
,
.
