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.![\[
f(x) = \sqrt{1 - x^a}, \quad a \geq 0
\]](http://latex.artofproblemsolving.com/e/f/1/ef14689f1894683256f9a0155037788366496929.png)
![\( x \in [0,1] \)](http://latex.artofproblemsolving.com/7/6/d/76d950c78d04a107e9fcb49b26ab353531c93b92.png)
.
, the function appears to be concave. However, I am uncertain about the behavior at the endpoints. and understanding the behavior at the endpoints? Any help would be greatly appreciated!
such that 
.
that divides 
for the positive integer 
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of positive integers such that 
in all of this cases :
of positive integers such that : 
3-Every prime number less than 
can be represented in the form 
by taking an Appropriate pair cannot be represented in the form where 
and are positive integers
. The number 
is the least prime number that cannot be represented as a sum or an difference of a power of and a power of . Prove that the number cannot be represented in any of the forms 
, 
, where and are positive integers
be real positive numbers such that: 
. Prove that 
+1 w
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
is a power series for which each coefficient 
is 
or 
. Show that if 
, then 
must be irrational.
.
s.t. 
.
, where ![$K=\mathbb{Q}[\cos(\dfrac{2}{3}\arctan(3\sqrt{3}))]$](http://latex.artofproblemsolving.com/a/2/e/a2efd64f27687f9ba5239cc4e0ba49d469d1c8c9.png)
.
.
and not its conjugate transpose.
.
.
; if the answer is yes, then give a method to calculate at least one of its roots.
. s.t. ., where .. and not its conjugate transpose.
is invertible we can write 
for some 
. Then 
hence is an orthogonal matrix.
.
has distinct eigenvalues which one can easily check.
of eigenvectors and the spaces 
are invariant under . Since is orthogonal we have 
and we find 8 solutions.
for some constants 
.
and that 
are obvious solutions.
and using this one can express 
on the basis 
. Then rewrites as 
.
where 
is the eigenvalue corresponding to 
.
. For each choice of signs we get one solution 
.
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
![\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]](http://latex.artofproblemsolving.com/9/5/9/9591f7c842d1b6afcba11b15b3ea3f9e01bc937a.png)
![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
