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.
such that 
for all 
.
such that the equation
such that 
. take a random 
on the altitude from 
to 
.
intersects second time in 
. Take 
such that it's on the segment and 
and 
.The second intersection of 
and circle 
is 
, (
) and the second intersection of 
and circle 
is 
,(
).The tangent from 
to the circle intersects the altitude from at 
.
is tangent to circle 
.
are 
and 
Find the product of all possible values of 
be a set consisting of 
positive integers, none of which is a sum of two other distinct members of . Prove that the elements of may be ordered as 
so that 
does not divide 
for all 
.
be an acute triangle with 
and 
. Let be the midpoint of the side 
. The circumcircle of triangle 
intersects the side again at 
and the circumcircle of triangle 
intersects the side again at 
. The point lies on the perpendicular bisector of the segment so that the points and lie on the same side of 
and 
. Prove that the circumcircles of triangles 
and 
are tangent.
![\[ p = \frac{2ab}{a + b} \]](http://latex.artofproblemsolving.com/0/f/4/0f4297ec59fe97c92fda75af4cc835be6b9a2507.png)
, ![\[ p, a, b \in \mathbb{C} \]](http://latex.artofproblemsolving.com/b/4/4/b4499970ac49d770a810d7345fedef7773cd03f5.png)
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
has rational sides and equal angles. Show that it is regular.
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 
.
and 
.
Prove that
Prove that
Let 
Prove that
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. 