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.
and 
for 
. Let 
be a positive integer. Suppose that for every positive integer 
there exists a positive integer 
such that 
. Must be a Fibonacci number?
satisfying following conditions.
for all 
.
with 
, equality 
holds.
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
. The line 
and the circle of radius centered at 
cross at 
and 
, respectively. Prove that , the circle on diameter 
and circle 
are concurrent.
+1 w
of functions from the set of real numbers to itself that satisfy ![\[g(f(x+y)) = f(x) + (2x + y)g(y)\]](http://latex.artofproblemsolving.com/0/a/3/0a3bc25299a0bcbd3f3602a0e6881d68fce750c5.png)
for all real numbers 
and 
.
such that for all 
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call to be a grandson of if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
of primes (greater than 5) such that : 
of positive integers such that:
.Prove that: 
has incentre 
and incircle touching 
at 
. Let 
be the antipode of in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
and 
intersect at points and 
. A line through intersects and at points 
and , respectively. Line 
intersects at point 
, and line 
intersects at point 
. If is the circumcenter of 
, prove that 
.
be such that![\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]](http://latex.artofproblemsolving.com/1/5/e/15e5bf0cc14794e7028dbfe4a8754f6803e850ea.png)
and 
. Find 
.
denote the set of positive real numbers. Find all functions 
such that for each 
, there is exactly one 
satisfying 
. It is easy to see that for any prime 
, we must have 
. Hence we get 
So we are done by discrete continuity if 
and 
for some 
. So we have two cases left:
for all .
. Let be the smallest non-negative integer such that 
Hence we have 
. Call this common value 
. Note that 
.
for 
. Also by our assumption 
. So 
form a set of at least 
consecutive integers, and hence every prime not exceeding divides at least one of those terms. But 
for all 
is divisible by every prime not exceeding 
, which is absurd because 
and 
. 
for all sufficiently large .
for all . Call a prime 
new if 
but 
for some 
. Since 
is unbounded and 
is bounded, the sequence 
must be unbounded. Therefore there must be infinitely many new primes. Choose any new prime 
, and let 
such that 
but 
. Contradiction! 
,
denote the product of distinct prime factors of 
.
,
for 
,
with 
.
,
?