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.
, there exists some positive integer 
that makes the following equation have no integer root 
.
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
such that the equality 
holds for all pairs of real numbers 
.
with 
.![\[
(x^{2}-yz)(y^{2}-zx)(z^{2}-xy).
\]](http://latex.artofproblemsolving.com/a/8/7/a87fae27584012a867bc0a90190c0317f9f7ab07.png)
be a positive integer. A positive integer is called a benefactor of if the positive divisors of can be partitioned into two sets 
and 
such that is equal to the sum of elements in minus the sum of the elements in . Note that or could be empty, and that the sum of the elements of the empty set is 
.
is a benefactor of 
because 
. has at least 
benefactors.
satisfying the following.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
such that for any positive real numbers 
, which are the sides of a triangle, the following inequality holds:![\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]](http://latex.artofproblemsolving.com/9/1/c/91c8a4af2bd2f3107c8921d2e06f8a09dedd0164.png)
. Let the perpendicular line from to 
meets 
at points 
, respectively, and the foot from to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at 
.
is tangent to 
.
be the midpoint of 
, 
be the midpoint of 
, let 
be where the perpendicular bisector of 
intersects . Note that
Therefore 
is cyclic. Observe as 
. Converse of Reim with the previous two facts gives 
cyclic. Therefore:
Therefore 
is equilateral.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
and 
: Prove that :
,
such that 
is on the same side of 
as 
and 
.
intersects 
at 
.
on 
such that 
is equilateral.
