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.
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)
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.
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points and are taken so that 
and 
. Prove that 
.
and points and on sides and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
with
such that 
divides 
for all positive integers and 
with 
.
such that 
Follows for all reals .
be a positive integer, and let 
denote the sum of the positive divisors of 
. Prove that the 
smallest positive integer relatively prime to is at least , and determine for which equality holds. to be 
-smooth if all of its prime factors are 
. Let 
denote the number of -smooth integers upto 
. Let 
be the least quadratic non-residue mod 
. Show that there are at least quadratic residues mod up to .
is a non-negative arithmetic function and
is convergent for some 
, then prove that
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
, let denote the smallest possible value of where are sets such that and whenever . Determine for each positive integer .
for 
.
if you consider this case.![$A_1,A_2\subset [1,2]$](http://latex.artofproblemsolving.com/f/9/e/f9ef3e6a63ae0c90866778c950a2f620a5581396.png)
is good . Then the 
are distinct, hence one of them equals 2. Contradiction. With 
we have 
.![$A_1,A_2,A_3\subset [1,4]$](http://latex.artofproblemsolving.com/5/b/2/5b264de1077e98d22ad161b90d443e6b6cb17696.png)
is good. Then the 
are distinct, hence wlog 
and 
with cardinalities 2,3. Hence one of them is 
, contradiction. With 
we see that 
.![$A_1,\cdots, A_n\subset [1,n+2]$](http://latex.artofproblemsolving.com/7/4/6/746e7ea19cfc02b32f40365d3a360564055e4ca8.png)
are good with cardinalities 
. Then we can construct the same for 
. Proof: let ![$B_k:=A_k\cup \{n+3\}, B_{n+1}:=[1,n+2],B_{n+2}=\{n+4\}$](http://latex.artofproblemsolving.com/f/5/5/f558bc8399b8d728f3b31dfd2d1fd3544643cbb5.png)
.
we find that 
for all 
. and 
. Then also 
.![$A_1,\cdots,A_n\subset [1,n+1]$](http://latex.artofproblemsolving.com/f/9/4/f947237f61f87303557379cc82df04f0accc17b8.png)
is good. The ![$|A_k|\in [1,n]$](http://latex.artofproblemsolving.com/8/2/b/82b387672583c83a13aa0c0e670109816a6d49ca.png)
are distinct hence wlog 
. Then ![$A_1,\cdots, A_{n-1}\subset [1,n]$](http://latex.artofproblemsolving.com/c/7/a/c7ab2f096ff0265b7ae4db392a4d1aff59a641ea.png)
are good. for some . Then 
and we can use Lemma 2 to obtain etc until 
. Contradiction, hence for .
for directly: suppose that is good. Then since 
and 
for 
and , the sizes of 
are 
. Hence we can assume WLOG . Then are good of sizes 
. WLOG 
. But then 
(here we use the assumption ), contradiction.
is not true when 
. The cardinalities are distinct, however, and that's why we have for the cardinalities. The rest of your argument is fine.
?
.