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such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
is a multiple of 
, we can write 
for some positive integer . This implies
is a multiple of 
, we have 
. By the divisor bounding inequality, we get
must be one of the numbers 
up to the largest multiple of less than or equal to 
.
..
for some positive integer 
. Since , we can write
for some positive integer 
. Expanding the right side, we get a polynomial in with leading coefficient 
. Since the left side has leading coefficient 
, we must have 
. This is impossible since and 
are both positive integers greater than 1.
. Since , we can write 
for some positive integer 
. By the "Product of Two Numbers is a Square" technique, there exist positive integers 
with 
such that
, we get 
, or 
. By the Four Numbers Theorem, there exist positive integers 
such that
, we must have 
. Therefore, 
and 
. This implies 
and 
. Writing 
and 
, we get
, we get
is divisible by 
. However, since 
, we have 
, a contradiction.
and 
for any positive integer .
, which always works.
also always works.
.
with 
.
Now let 
.
so 
But then 
contradiction. So the only solutions are 
where 
. 
and 
. If 
, can be any positive integer (apart from ) so suppose that 
.
where is a positive integer. Now look at this equation mod and keep in mind that 
. We get 
so 
and 
. Now we have 
we get a solution so suppose that 
. We clearly have 
and 
. Now we have 
. After expanding, this implies 
, we are done, so assume not. Let 
, we have 
, 
, so 
. Since 
, 
.
, we have 
, so let 
, 
. 
, a contradiction.
, then 
, unless 
, which works, so we're done.
works, so assume 
.
and 
where 
are integers and is maximized. Also note that 
because 
. We see that we can let 
and 
where 
is an integer. Since , this means that 
. Plugging this in gives 
so 
so 
. Plugging in 
gives 
, which works. When 
, the LHS is bigger than the RHS.
and 
.
Either or 
. For case, 
obviously works
for a 
or in other words 
. We also have 
for 
. Subtracting these two gives us 
for some 
But we have 
So this case can't be true
This obviously works.
But we have , so this case can't be true,too
are positive integers.
for some positive integer .
then it is obvious that the conditions won't fullfill,it is easy to see that it works for all positive integers 
for all positive integers .
for some 
Hence, 
Hence, we conclude 
If 
then 
which is clearly absurd since our second condition implies 
Thus, 
and so 
is the only solution.
of positive integers such that 
If 
then we must have 
as the only solution. Note that if 
from the second condition. Suppose the power of a prime 
divides 
Then, 
divides 
This implies that 
divides 
divides 
Thus, every power of a prime that divides 
so 
But since we know 
we must have 
which always works.
give us 
?
for a positive integer 
Since 
and 
.
.
and 
and 
,
.
.
and 
.
but 
so 
.
so 
so taking modulo 
,
)
so 
so 
but 
works for any 
,
are coprime. Clearly 
.
and thus 
for some 
,
.
and therefore 
.
,
which holds only when 
,
and thus 
is the other solution family; concluding the problem.
(i.e. we clearly have 
and 
,
which means that for any 
we get that 
and using those along with the fact that all 
'
,
and 
,
(using euclidean alg) we use the second condition in this way, 
means 
which means that 
so 
which means that 
which gives 
but now this means that 
so 
which means that 
thus we are done 
,
for some 
and 
.
divides 
.
are coprime to 
.
.
.
,
and 
.
.
.
as well by noting 
or 
and bash cases.
.
,
so assume all 
.
and let 
since 
,
and so 
.
,
.
so 
.
for 
,
.
and 
.
,
.
.
Let 
Case 1: 
.
,
where 
.
which contradict 
Since 
Since 
,
so 
which give 
work.
,
and 
with 
.
since 
.
since 
.
or 
.
then
but because 
works.
for all positive integers 
(
.
for some 
.
Also we have 
So 
So 
Thus 
for all positive integers 
works.
then we have that 
and hence we get that our solutions are 
![\[0\equiv a^3-1 = kb^2-1\equiv k-1 \pmod{a-1}.\]](http://latex.artofproblemsolving.com/3/5/4/3541e7d8ab3bb04b9ce9875e1bbfcfe4283b61b3.png)
However, 
This forces 
in which case 
or 
Which is obviously impossible. Convsersely, We can easily verify that 
.
.
Thus, 
(if 
i.e. 
and 
.
which we can rewrite as 
.
.
implies 
.![\[\frac{b^2}{d^2} \leq d \implies a^2 \leq b^2 \leq d^3 \leq (k - 1)^3 \implies a^{\frac{2}{3}} + 1 \leq k.\]](http://latex.artofproblemsolving.com/2/2/0/220611cbf39624e602e0a89d55c497724f93f27f.png)
Plugging this back in, we receive 
which implies 
,
and 
for a positive integer 
.
from 
.
,
so 
.
or 
. 
gives 
,
,
and 
,
.
.
.
{Case 1: 
,
:
,
and 
.
and 
.
.
.
,
and 