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.
for which there exists an infinite subset 
of the positive integers such that: for any pairwise distinct positive integers 
, the sum 
and the product 
are both square-free.
be a finite set of real numbers, 
be a real number, and 
be 2025 non-zero real numbers. Define![\[A =
\left\{
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d
\right\},\]](http://latex.artofproblemsolving.com/2/e/1/2e18b6d25bfaa3e3d52f7efe5e4cba7a9cb87db9.png)
![\[B =
\left\{
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0
\right\},\]](http://latex.artofproblemsolving.com/4/1/f/41f6b126cc3cf24757b37147906c767d49a0fdd9.png)
![\[C =
\left\{
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0
\right\}.\]](http://latex.artofproblemsolving.com/d/e/6/de6586b88004e3f5aa7774525b00ee975c072ff9.png)
Show that 
.
and 
satisfies the following recursion formulas: 
, 
and 
. Show that 
, for 
.
be a finite, nonempty set of points in the plane such that, for every point 
in , there exist points 
in (distinct from ) such that 
. What is the smallest possible number of points in ? and are such that, for every positive integer 
,![\[ a_n > 0,\qquad\ b_n>0,\qquad\ a_{n+1}=a_n+\dfrac{1}{b_n},\qquad\ b_{n+1} = b_n+\dfrac{1}{a_n}. \]](http://latex.artofproblemsolving.com/4/f/c/4fc9fd1cb31a25378268f20e019e639667e52910.png)
Prove that 
.
and centroid 
. 
is the reflection of over , similarily define 
and 
. over , it is the same as reflecting over 
over and over , 
over and over .?
be an isosceles triangle with 
, and let 
be a point inside side 
such that 
. Let 
and 
be two points inside sides 
and 
, respectively, such that 
. Let the perpendicular bisector of 
meet line segment 
at 
, and let the circumcircles of triangles and 
meet again at point 
, different from ., , are collinear. Prove that 
.
with 
with the property that![\[\varphi(a^m-1)|a^{\varphi(m)}-1 ?\]](http://latex.artofproblemsolving.com/e/0/7/e07ae15e0267f4725151c919b47aaac7b950351d.png)
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
, 
and 
. The interior bisector of the angle 
intersects the side at the point 
. Calculate the length of the line segment 
.
, 
, 
. If the altitude to the side whose length is divide the side into two line segments prove that their difference is 4. such that
, 
and 
. Prove that 
.
with real coefficients satisfying
for all real numbers 
with integer coefficients such that, for any positive integer , the equation 
has an integer root.
is a positive integer,prove that 
holds
, so, 
, thus, holds).
are positive real numbers then prove that 
.
are positive real numbers 
. Let 
, 
, 
. Then the inequality becomes 
. Integrating both sides of this from 
to 
, with respect to 
, gives 
hence 
.
are positive real numbers are positive real numbers then prove that is a positive integer,prove that are three positive real numbers in harmonic progression .Prove that 
are three positive real numbers in harmonic progression .Prove that 
and 
Prove that
are positive real numbers 
and prove 
obviously, and also 
.
,
,
,
,
from an arithmetic progression. Let 
be the common difference of this sequence. We get 
and 
.
and 
respectively and combining fractions we get that we want to prove
so it suffices to show 
.
are positive, so we get 
.
,
,
,
to get this estimate.