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.
is strictly increasing : 
such that:
, let 
be the set of positive integers that do not exceed and are coprime to . Define 
as the smallest positive integer that allows to be partitioned into disjoint subsets, each forming an arithmetic progression.
satisfying 
, 
, and 
.
be a real numbers such that 
Prove that
Where 
be natural number and 
be the set of distinct natural numbers. Define function 
with 
. Prove that have at least distinct values.
and 
such that
is a prime number.
be the midpoint of the side 
of triangle 
. The bisector of the exterior angle of point 
intersects the side in 
. Let the circumcircle of triangle 
intersect the lines 
and 
in 
and 
respectively. If the midpoint of 
is 
, prove that 
.
and satisfying the equation
?
is not a perfect square, then there exists an odd prime s.t. is a quadratic non-residue mod .
be the orthocenter of the triangle . Let and be the midpoints of the sides and , respectively. Assume that lies inside the quadrilateral 
and that the circumcircles of triangles 
and 
are tangent to each other. The line through parallel to intersects the circumcircles of the triangles and in the points 
and 
, respectively. Let be the intersection point of 
and 
and let 
be the incenter of triangle 
. Prove that 
.
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
has always integral solutions for 
and 
whenever 
is a positive integer.
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
defined for 
being an interval (connected subset of the real line) and 
not being in this image, it follows that we may assume that 
for all . But then 
for all 
and so 
is a minimum point of 
. Of course, this can be written in 11-th grade vocabulary. 
has the same sign for all 
.Can you be give me more detailes please 
?The official solution is based on the fact that the function 
is injective ,hence monotonic.
such that is a connected subset of the plane and the function 
is continuous on this domain, thus its image is a connected subset of the line, thus an interval. I agree however that this is not a solution of an 11-th grade student, but that's how life is. I will not be amazed if in a few years I see complex analysis, Lebesgue integration and other such stuff at RMO. It's quite à pity, it gives huges advantages to some people. 
should be defined on the set of pairs such that 
to ensure that its domain is connected. Of course, all the rest works with this modification, I don't know how I could write such a stupid thing. 
, then 
is strictly concave up or down on some neighbourhood of the point , thus one can draw a close enough parallel to the line tangent at to the function's graph that intersects the graph in two points 
and 
. The slope of that tangent would be 
, equal to the slope of the tangent at , that is,.
is connected, but why 
is connected ? has clearly two connected components. However with this spirit we can also solve the problem, extend 
to whole plane defining 
and now 
is continuous on whole plane and do similar thing.
has the same sign for all .Can you be give me more detailes please ?The official solution is based on the fact that the function is injective ,hence monotonic.
there exists an with the desired property. So you get 
only for values in some set , which has to be dense in the reals, but need not be an interval, so the IVP you use later on in your proof does not give a contradiction. (See the solutions above for proofs that avoid this problem.)
is continuous on 
and that is true because is twice differentiable.
. In the first paragraph you show that for all 
has to contain some 
(which means that is a dense subset of the real numbers). This part of your proof is fine. But in order to make the proof of the green claim work you need to show that contains all real numbers. This is missing in your proof (if I interpret you proof correctly).
for all 
and now make 
close to 