AoPS Academy
. How many complex numbers 
are there such that 
and 
is an integer.
and 
be points on a plane such that 
, where 
is a positive integer. Let 
be the set of all points 
such that 
, where 
is a real number. The path that traces is continuous, and the value of is minimized. Prove that is rational for all positive integers 
be complex numbers with 
. Find the greatest possible value of the expression 
digit numbers are there
? and 
?
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solve
integer matrices having the following properties:
the entries of are (positive) prime numbers,
there exists a integer matrix such that 
and the determinant of is the square of a prime number.
such that the equation in the unknown 
:
admits 
roots: 
s.t. 
.
and 
Prove that 
and Prove that
![\[
\begin{cases}
y = \cos \alpha \\
x = \sin \alpha \cos \beta, \text{ with } \alpha, \beta \in \left[0, \dfrac{\pi}{2} \right] \\
z = \sin \alpha \sin \beta
\end{cases}
\]](http://latex.artofproblemsolving.com/2/6/7/2676398fe8bfc519baf72b54abef7d9b4c25606f.png)
Then 
![\( \alpha, \beta \in \left[0, \dfrac{\pi}{2} \right] \)](http://latex.artofproblemsolving.com/d/9/e/d9ed45f138b74a247de2b7167192da6120ab610c.png)
, we have 
![\[
Q \leq \dfrac{\sqrt{2}}{2} \sin 2\alpha + \dfrac{ 1 - \cos 2\alpha}{2} = \dfrac{1}{2} + \dfrac{1}{2} (\sqrt{2} \sin 2\alpha - \cos 2\alpha) \le \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\sqrt{ \sin^2 2\alpha + \cos^2 2\alpha} = \dfrac{1+\sqrt{3}}{2}
\]](http://latex.artofproblemsolving.com/1/a/2/1a2ffb2ae68498b4b644d3a17640693457837a2b.png)
Equality holds when![\[
\dfrac{\sqrt{2}}{2} \sin 2\alpha - \dfrac{1}{2} \cos 2\alpha = \dfrac{\sqrt{3}}{2} \\
\Leftrightarrow \dfrac{\sqrt{2}}{2} \sin 2\alpha - \dfrac{1}{2} \cos 2\alpha = \dfrac{\sqrt{3}}{2}
\Rightarrow \sin 2\alpha = \dfrac{\sqrt{6}}{3}, \cos 2\alpha = \dfrac{1}{3}
\]](http://latex.artofproblemsolving.com/f/3/8/f38d5f3386b6022aa0c2734ee622c44dc5320f56.png)
So we find![\[
\begin{cases}
\sin \alpha = \sqrt{\dfrac{3 + \sqrt{3}}{6}}, \quad \Rightarrow y = \sqrt{\dfrac{3 - \sqrt{3}}{6}} \\
\cos \alpha = \sqrt{\dfrac{3 - \sqrt{3}}{6}}, \quad \Rightarrow x = z = \sqrt{\dfrac{3 + \sqrt{3}}{12}}
\end{cases}
\]](http://latex.artofproblemsolving.com/c/6/6/c66335412926c36b5b37c809c57cbc69be629b30.png)
Thus, the maximum value is 
when![\[
y = \sqrt{\dfrac{3 - \sqrt{3}}{6}}, \quad x = z = \sqrt{\dfrac{3 + \sqrt{3}}{12}}
\]](http://latex.artofproblemsolving.com/2/a/a/2aac73ae26d015362ba6f34b6cc360c3660a0ceb.png)
and 
Prove that
![\[\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)}.\]](http://latex.artofproblemsolving.com/7/d/6/7d6247fbcfd6eac156ce774f727d7e95c470de52.png)
![$\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{lcm(1,2,3,...,n)}=e$](http://latex.artofproblemsolving.com/a/1/3/a139e5cb990f3386de109c701f8edc71b4054712.png)
is wrong, must be 
!
Prove that 
be real numbers satisfying 
![\[ xy + yz +2xz \le \dfrac{1+\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/a/7/2/a72a5d2df62fa5adbdf58d7b0bde43bfc548a797.png)