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.
be non-empty subsets of the positive integers, and let 
denote their union. Prove that![\[\sum_{i=1}^m \sum_{(a,b)\in H_i^2}\gcd(a,b)\ge\frac1m \sum_{(a,b)\in S^2}\gcd(a,b).\]](http://latex.artofproblemsolving.com/1/c/d/1cdf561d3af92f524628e0724e0c5ee24d8d6a09.png)
Proposed by Dávid Matolcsi, Berkeley
. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
and 
Find the value of 
, such that 
be an infinite sequence of real numbers, for which there exists a real number 
with 
for all 
, such that ![\[\left\lvert a_i-a_j \right\rvert\geq \frac{1}{i+j} \quad \text{for all }i,\ j \text{ with } i \neq j. \]](http://latex.artofproblemsolving.com/a/4/d/a4d87553590a450365b6525ca62a5d0416fe41cc.png)
Prove that 
.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
which satisfy the following set of equations
inscribed in 
with 
being the midpoint of 
. The tangents at 
of intersect at 
. Let 
be the projection of 
onto 
. On the perpendicular bisector of , take a point 
that is not on and different from M. Circle 
intersects 
at 
. Lines 
and 
intersect at 
. Prove that 
is an isosceles triangle.
, and 
be positive real numbers, such that 
. Prove the inequality![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
be a non-constant monic polynomial with non-negative integer coefficients such that 
. Let 
be a sequence such that 
and 
. Prove that for every , there exists a prime 
such that 
but 
.
and integers 
, 
. Since 
, we get 
Suppose that for some 
, and each dividing 
, there exists such that 
. Then from the claim we get 
But we have 
which is a contradiction., there exists a prime such that but .
.
is injective modulo 
for any 
prime (e.g., see my solution to a 1999 Balkan problem).
such that 
. Take such a pair 
with the smallest coordinate sum and assume 
. Then, 
forces 
, contradicting with minimality of . So, 
. But then, there is a 
such that 
, forcing 
.
Prove that
Prove that
satisfying
Prove
and 
for any prime 
Indeed, we proceed using an induction.
,
for some 
.
where there are 
to finish the induction.
for any prime 
:
for some 
to finish the induction.
be the set of prime divisors of 
.
is infinite, as required. 