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.
denotes the least natural such that
Find all naturals 
such that 
.
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained?
, 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.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
and 
: Prove that :
,such that 
Prove that : 
Where 
is a positive real number solution of equation : 
![\[
u=1+\sqrt{1+ka},\quad v=1+\sqrt{1+kb},\quad w=1+\sqrt{1+kc},
\]](http://latex.artofproblemsolving.com/1/3/e/13e016c14c1ba8467ac11bcd217e114fcd067a8d.png)
and note that![\[
a=\frac{(u-1)^2-1}{k}=\frac{u^2\Bigl(1-\frac 2u\Bigr)}{k}= \frac{u^2\Bigl(\frac{2}{v}+\frac{2}{w}\Bigr)}{k}.
\]](http://latex.artofproblemsolving.com/0/2/f/02f24b6ce57cca2486ab567b3b0f9aa7b4573902.png)
![\[
xyz=\sum ax^2=\frac{1}{k}\sum \Bigl(\frac{2}{v}+\frac{2}{w}\Bigr)(ux)^2=\frac{1}{k}\sum \frac{2}{u}\Bigl[(vy)^2+(wz)^2\Bigr]\ge \frac{1}{k}\sum\frac{2}{u}\cdot2vywz.
\]](http://latex.artofproblemsolving.com/3/d/2/3d22da4a6db42ca431539f792aafd26f7e1a2d9d.png)
yields![\[
1\ge\frac{uvw}{k}\sum \frac{4}{u^2 x}.
\]](http://latex.artofproblemsolving.com/7/8/6/786bfb004ad038ed6fe576141a753e380792ac3f.png)
![\[
\sum x\ge \frac{uvw}{k}\Biggl(\sum \frac{4}{u^2 x}\Biggr)\Biggl(\sum x\Biggr)\ge \frac{uvw}{k}\Biggl(\sum \frac{2}{u}\Biggr)^2=\frac{uvw}{k}.
\]](http://latex.artofproblemsolving.com/e/b/d/ebd9b73d8809fa523f11604c2e7ed47b742b01a8.png)