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
satisfies the given constraint then so does 
, and we can apply the same argument to that sequence.
if and only if 
have different sign, so for any fixed value of 
, 
should have fixed sign (positive, negative, or zero). Clearly if 
then 
for all 
as well, else take 
and 
, and then the desired claim is true. Thus assume that no terms in the sequence are 
. Then if 
then 
is forced by comparing 
with 
, and by induction eventually either every term is at most or exactly other term is at most . We only care about the latter scenario, in which case we care only about the subsequence 
of terms greater than , which "behaves" the same way as a complete sequence in itself would. Thus assume that every term is greater than .
for each 
, so each 
is a positive, and it suffices to show is bounded below by a positive real. The condition (after reciprocating both sides) then becomes 
. This is then a linear recurrence whose solutions (over 
) are 
. But if 
then we eventually have 
which is forbidden, so 
and the are bounded below by 
. 
be a sequence of positive numbers satisfying, for any positive integers 
such that 
,
Show that there exist positive numbers 
so that 
for any positive integer 