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.
. Does there exist a function $f: \mathbb{N}__0 \to \mathbb{N}_0$ such that:![\[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0 \]](http://latex.artofproblemsolving.com/7/a/d/7ad2ffcca35dbc32a84a35a86fefa260fe7c5a85.png)
and 
, 
?
and all positive integers 
and 
satisfying 
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers 
such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
be an odd prime number and a sequence 
satisfy 
and 
Prove that always exists positive integer satisfying for all positive integers , then 
.
is "not congruence"
and 
Prove that
Solution:
and 
then 
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 
.
, whose center is O, we have OB = OP.
.
, it thus follows that this point N is the center of the circle . Since the line PT is tangent to this circle at the point T, we thus have 
. Hence, < PTN = 90º, so the triangle PTN is right-angled at T, and hence, by Pythagoras, 
.
. On the other hand, since the point T lies on the circle , whose center is N and whose radius is 
(since OM is a diameter of this circle), we have 
. with its center O and its chord AB, we obtain that the point O lies on the perpendicular bisector of the segment AB; in other words, 
, since M is the midpoint of the segment AB. Thus, < OMB = 90º, so that the triangle OMB is right-angled at M, and thus Pythagoras yields 
. Now, remember that OB = OP; also, since M is the midpoint of the segment AB, we have 
. Thus, 
.
,
, and the problem is solved.
with centre , let 
be a chord that is not a diameter. Let 
be the midpoint of this chord . Take a point 
on the circle with 
as diameter. Let the tangent to at meet at 
. Show that .
be the midpoint of . is the origin. Let 
, 
, 
, 
. We can easily see that 
and 
.
is 1, i.e. 
. Thus, 
gives us
Now, it suffices to show 
, i.e.
which is clearly true as 
is a right triangle.
and 
.
,
.
that satisfies:
and 
.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
Prove that
Prove that
be the midpoint of 
and we have 
and we have know that 
and 
and with these we R done.(there is only calculate) 
to be the midpoint of 
.