[/quote]
,
and 
for 
.
be a positive integer. Suppose that for every positive integer 
there exists a positive integer 
such that 
.
satisfying following conditions.
for all 
.
with 
,
holds.
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
.
and the circle of radius 
cross 
and 
,
and circle 
are concurrent.
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 
.
be positive integers. We call 
of primes (greater than 5) such that : 
of positive integers such that:
.
and incircle touching 
at 
.
be the antipode of 
is chosen on the internal angle bisector of 
such that 
.
be the midpoint of arc 
,
be the midpoint of 
.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
and 
intersect at points 
.
and 
intersects 
,
intersects 
.
,
.
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 
.
.
denote the set of positive real numbers. Find all functions 
such that for each 
,
satisfying 
.
Y by megarnie, Adventure10
Let 
and 
be distinct positive integers. The following infinite process takes place on an initially empty board.
, operation 
will be performed only finitely many times.
Proposed by Serbia.
and 
be distinct positive integers. The following infinite process takes place on an initially empty board.- If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by and the other by .
- If no such pair exists, we write two times the number
.
, operation 
will be performed only finitely many times.Proposed by Serbia.
This post has been edited 4 times. Last edited by Tsukuyomi, Jul 17, 2019, 12:56 PM
such that we have to use operation 
infinitely many time. We called the final position of each time as 
(Final position before we use operation 2)
.
and 
.
to count the number of each number appear in the process after we introduce 
doesn't depend on the order too. (The proof is easy induction.) So, if, in some case, we can make infinitely many numbers that 
then we are done.
satisfy the Claim. The proof is bashing everything out. (The actual proof is too messy and I'm too lazy to show that.)
and 
.
and 
are on the board, and we add 
,
will appear on the board twice for all nonnegative 
,
.
then 
and 
will both appear twice by the induction hypothesis, and these will each "spawn" a copy of 
and 
,
will appear twice, and so we get at least one copy of 
.
.
where 
and 
.
,
appears on the board twice, by the induction hypothesis. This number will hence "spawn" another copy of 
.
(because 
,
will appear in the terminating state of the board if we are on a round whose binary representation has a 
in the 
,
.
(in binary, where there are 
,
nonnegative integers and 
.
,
,
denotes the real root of 
greater than 1.
be the multiset of integers currently written on the board. Define the quantity![\[S=\sum_{n\in\mathcal{B}}f(n). \]](http://latex.artofproblemsolving.com/d/1/f/d1f49d2fc3cf8e45c9f61b39069ce62bbd6112d8.png)
We claim that 
is preserved under type (i) operations. It suffices to show that 
,
,
.
and 
(by rearrangements of 
)
.![\[S<\sum_{x=0}^{\infty}\sum_{y=0}^{a-1}\varphi^{2y-x}=\left(\sum_{x=0}^{\infty}\varphi^{-x}\right)\left(\sum_{y=0}^{a-1}\varphi^{2y}\right), \]](http://latex.artofproblemsolving.com/5/0/e/50e3a80294f35ad2c3031177f6720036690ded25.png)
which is finite since the left factor is a decreasing geometric series and the right factor is finite.
![\[2f(n) = 2\varphi^{2y-x}\]](http://latex.artofproblemsolving.com/f/a/f/faf56742ab8395ff3af1a44d0ec6f273e7393f04.png)
![\[f(n+a)+f(n+b) = \varphi^{2y-x+2} + \varphi^{2y-x-1} = \varphi^{2y-x}(\varphi^2 + \varphi^{-1}) = 2\varphi^{2y-x+1}\]](http://latex.artofproblemsolving.com/3/3/f/33f6e2b0d2b40e8cdc58d6dd8d27beae22dd8cca.png)
and these two are not equal? I had a similar idea of firing points, with 
where 
.
and 
when 
There is no reason to care whether it could be achieved by the condition that enforces the second rule on using operation. We could swap in any way we want.
that is the function which counts the number of times 
for 
iff 
for some constant 
.
with 
we needed operation (ii) an infinite number of times.
s
.
.
.
to be a sufficiently large integer, which is possible because we assumed we applied (ii) an infinite number of times, we can get, by rearranging the appropriate steps, some multiset of numbers of the form 
.
,
.
.
this multiset. But then if you apply the move to 
,
until we get to an "end position" where we must apply a (ii) move. If we apply an arbitrary sequence 
of type (i) moves on the same starting position, we will get to the same end position.
consists of replacing 
with 
then take the first move 
consisting of replacing 
(there must exist one, otherwise 
Note that 1) all moves are still valid since the 
Then apply the inductive hypothesis. 
for an integer 
.
times of operation (ii), I claim that this will go on forever. Let 
,
for 
,
,
things, then each of these numbers will get written at least twice, so we are done by the above paragraph. 
and define 
Inductively. If we have 
on the board, we can add 
to the board. Adding another 
on the board. Keep adding 
and 
will be on the board. Now, proceed to pair numbers arbitrarily until we reach all numbers being distinct. To show that we will use (ii) finitely often, it is sufficient to show there are finitely many 
'
,
,
.
are distinct. So, ![\[ k(b-a) \geq 2^k \]](http://latex.artofproblemsolving.com/e/5/a/e5a6b93b3b2211c7ab4d8708306b420c1c325621.png)
Which is not true for sufficiently large 
)
type 
)
type 
the base case 
is trivial. Notice that due to our assumption sequence 
must end at some point. Now suppose that 
replaces 
in particular the initial configuration of numbers contain two copies of 
so some move 
has to make replacing 
.
lead to the same final configuration as 
It suffices to apply inductive step for configuration appeared after 
of moves starting from an empty board, which contains 
zeroes will lead to the same final configuartion.
for 
with some sequence of moves type 
,
with 
is an integer, 
,
.
,
consecutive integers that are used during some round. Let 
be an integer such that on or before round 
,
.
and 
,
to have infinite occurences of 
.
between them.
.
.
points, and apply operation (i) any number of times. Then using the coordinates it is easy to construct the set in the claim via a sequence of right moves and up moves, so by the claim we are done.
to the first quadrant by a series of knight moves, which was used in a number of other solutions.
and 
.
in that order such that 
has at least 
,
can be written in this form. Thus if we make 
and at least 
for some large enough 
,
,
The configuration of the problem clearly prompts us to consider an alternate formulation of the problem. The following infinite process takes place on an initially empty coordinate grid.
,
represents a number at 
,
,
,
and 
,
,
,
and 
.
with 
.
and 
.
we assign a frequency of numbers 
,![\[(x,y),(x,y)\to (x+1,y),(x,y+1).\]](http://latex.artofproblemsolving.com/0/7/0/0702c2cf5500ae88799c8c7154d17a62becdcdb5.png)
If 
,![\[(x,y),(x,y)\to (x+1,y),(x+b,0).\]](http://latex.artofproblemsolving.com/6/5/0/650558d2b3cadce0c018bf05d0e7df9a0c1c9d9c.png)
Operation 
.
operations of 
are all 
for 
may only be incremented through the step![\[(r-1,0),(r-1,0)\to (r,0).\]](http://latex.artofproblemsolving.com/3/3/7/3370b57cf36a1d905df22be3dc57fc33dc9dc919.png)
Evidently we begin with a value of 
at 
times at 
.
are all 
t
consists of lattice points 
.
.
times. We claim that operation 
such that 
,
,
are linear combinations of 
and at least one copy of 
,
