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 a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be positive integers, where 
. It is given that there exist a positive integer 
such that
. Prove that the above equation is true for all positive integers 
. (Here 
is the sum of digits of 
taken in base 
).
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
, there exist pairwise different real numbers 
satisfying
be a positive integer, and let 
denote all positive divisors of , If the following conditions are satisfied:
.
satisfying the equation 
.
.
and 
but 
because 
. I mean 
and 
are perfect squares. 
. a should be 0 or 1 because of . We have two solution. These are $(a,b)=(0,0),(1,0)
be a triangle with 
. The point is the midpoint of the side 
. A circle with center at vertex 
intersects the line at points and 
. Prove that 
.
and 
Since 
, thus
We want to show that 
., 
and 
over angle bisector 
and obviously we see that goes to and let point 's reflection point be 
. Let 
be the intersection point of angle bisector and line .
, hence we only need to show that 
.
and since we reflected over the line 
, then 
and 
.
Hence, we are done yay.
: symmetrical point of C to bisector of 
such that 
where 
and 
Prove that
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
,
is the binomial coefficient 
choose 
.
