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.
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
, and 
for all integers 
. Prove that 
.
and 
Prove that
Let 
and 
Prove that
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
and 
intersect at 
. The common tangent line nearer to 
of the two circles touches 
at 
respectively. Let 
be the symmetric points of with respect to respectively. The circumcircle of triangle 
intersects circles and at points 
respectively which are distinct from . Prove that the circumradii of the triangles 
and 
are equal.
, 
. 
intersects 
at 
. A line 
cuts segment 
at 
respectively. If 
and 
, prove that the diameter of the circumcircle of 
equals 
.
are points on a circle 
, satisfying 
, 
. meets 
at 
. 
is on 
such that 
. Suppose 
intersects again at 
. 
is the reflection of 
wrt 
. Prove that 
lies on the same circle.
in its exterior
we draw the triangles 
so that 
, 
, 
.
and
and be on segment of an acute triangle 
such that 
and 
. Let and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
has a right angle at 
. Let be the point on line 
such that 
and 
lies between and 
. Point 
is chosen so that 
and is the bisector of 
. Point 
is chosen so that 
and 
is the bisector of 
. Let be the midpoint of . Let 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and a point such that 
and 
are tangent to the circumcircle of . Let 
be the reflection of over and 
be the reflection of over 
. If 
is the circumcenter of 
, help Oscar prove that 
is perpendicular to .
so 
but your result would be 
.
...
We know that 
and 
the sequence of power in 
is the power of first 
terms in 
and is the last 
terms in We claim that is strictly increasing and analogously is strictly decreasing
We just need to prove that 
so we are done. is odd so all of the power is even. It is east to see that 
and also 
So we claim that 
the first terms we have 
in the last terms such that 
we know that we have a multiple of 
in the last terms now take the greatest multiple of so we have 
now we just need to prove that 
and this is absolutely correct. Now 
take
are given such that 
.
in natural numbers m and n.
such that 
and 
Prove that there exists infinite number of 
is a square of an integer.
![\[a_1a_2\dots a_n=(n+1)^n \cdot n^{n-2} \cdot (n-1)^{n-4} \cdots 3^{4-n} \cdot 2^{2-n}.\]](http://latex.artofproblemsolving.com/3/4/b/34b3ef7e33326179775fcd6faf3a7eb982248ded.png)
Now for even 
is an odd square, then our number is a rational square.![\[(n+1)^{n-1} \cdot n^{n-2} \cdot (n-1)^{n-1} \cdots 3^{4-n} \cdot 2^{2-n}\]](http://latex.artofproblemsolving.com/b/4/f/b4fadf5df6b0769a5144d92457a13d2bc3941d50.png)
is always an integer when 
,
do the job.
is a prime, we need to check that![\[(n-1)v_p(n+1)+\sum_{k=1}^n (2k-n-2)v_p(k) \ge 0.\]](http://latex.artofproblemsolving.com/9/2/5/9252cb37fa03e1579e6e391baa08d598cbbc3f55.png)
Separating moreover the contributions from all powers of 
we have![\[(n-1)\chi_{q \mid n+1}+\sum_{k \le n, q \mid k} (2k-n-2) \ge 0.\]](http://latex.artofproblemsolving.com/b/f/3/bf36000e6d6823a65849f0030a16c4d5880239ab.png)
Computing the LHS, we get the equivalent![\[(n-1)\chi_{q \mid n+1}+q \cdot \left\lfloor \frac{n}{q}\right\rfloor \cdot \left(\left\lfloor \frac{n}{q}\right\rfloor+1\right) \ge (n+2) \cdot \left\lfloor \frac{n}{q}\right\rfloor\]](http://latex.artofproblemsolving.com/0/9/8/0986d4ffdfab4df2e859e2190053cdd39c838314.png)
Writing 
with 
,![\[(n-1)\chi_{r=q-1}+q \cdot a(a+1) \ge (aq+r+2) \cdot a\]](http://latex.artofproblemsolving.com/d/a/c/dac0edaaaf9a5e4f71924dde7e7fc83606293caa.png)
or equivalently![\[(n-1)\chi_{r=q-1} \ge a(r+2-q).\]](http://latex.artofproblemsolving.com/d/0/1/d016a3ba3833bb6b2cac008518a12ef03cdfa31e.png)
If 
,
and 
and if 
we need to check that 
which is obvious.
so we can see 
it would be sufficient , so we can say it is an square of an integer because 
