find the number of ordered real triples 
such that![\[\begin{aligned}
\begin{cases}
x+y^2+z^2=a,\\
x^2+y+z^2=a,\\
x^2+y^2+z=a.
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/9/8/6/986593653bf128bf7b3d497abed78e1459134c5f.png)
such that ![\[ \dfrac{a^2}{2ab^2-b^3+1} \]](http://latex.artofproblemsolving.com/0/d/2/0d2f10786d22af2ea94ebeed6b15e723f4042734.png)
is a positive integer.
?
.
s.t. 
.
,![$K=\mathbb{Q}[\sin^2(\dfrac{2}{3}\arctan(3\sqrt{3}))]$](http://latex.artofproblemsolving.com/5/b/8/5b87dd74c6659350c000071d5a9228f2a80b1466.png)
.
.
and not its conjugate transpose.
be a finite abelian group. There is a magic box 
.
contains only the group identity, 
.
be a positive real number. Divide the positive real axis into intervals 
,
,
,
,
,
satisfying the following differential equations:![\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]](http://latex.artofproblemsolving.com/2/6/8/268b7bc6150b13eed666d566b540df7e643d8344.png)
![\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]](http://latex.artofproblemsolving.com/b/4/6/b46d0d00b34cd3fb72ec310625384547b14871c5.png)
with the initial conditions:![\[
u(0) = 1, \quad u'(0) = 1,
\]](http://latex.artofproblemsolving.com/4/1/6/416754efbff23b7efbf02b32149b1e1fd5cb9905.png)
and the continuity conditions:![\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]](http://latex.artofproblemsolving.com/b/3/0/b309f7b8c70cb27683a96af8452ebf9e36f67a4c.png)
Show that![\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]](http://latex.artofproblemsolving.com/f/0/b/f0b9105523f0561e0a4d151a2695cbf6917e5cc7.png)
![$x\in [0,4]$](http://latex.artofproblemsolving.com/a/8/9/a89e045d4c66cdb788e64e613bd6c3c9b5931cc2.png)
at which the function 
takes its maximum value
be an algebraic extension of 
,
be an algebraic closure of 
is an embedding such that 
for all 
,
extends to an automorphism of 
is an embedding. Then, there exists an embedding 
that extends 
,
is an isomorphism.
and 
that extends 
is also an algebraically closed field.
,
be the field of algebraic numbers, 
.
defined by
Then, according to the exercise above, 
How should we interpret 
?
,
.
such that![\[\mathcal{E}(B \cup C) = B \cup C.\]](http://latex.artofproblemsolving.com/e/c/3/ec34115c9e6e169c19c739220065f647934a1a5f.png)
Y by Mango247
For every positive integer 
prove the inquality
(The integer part![$[x]$](//latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
of the number 
is the largest integer not exceeding . The fractional part of the number is a number 
such that ![$[x]+\{x\}=x$](//latex.artofproblemsolving.com/0/3/a/03ae0917760023ab705aee0ca548af5adc9ba945.png)
.)
A. Golovanov
prove the inquality
(The integer part
![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
of the number 
is the largest integer not exceeding . The fractional part of the number is a number 
such that ![$[x]+\{x\}=x$](http://latex.artofproblemsolving.com/0/3/a/03ae0917760023ab705aee0ca548af5adc9ba945.png)
.)A. Golovanov
This post has been edited 6 times. Last edited by Hopeooooo, Jul 25, 2024, 5:09 PM
Reason: 1
Reason: 1
where 
is the fractional part of 
.
.
.![\[
|2\sqrt{m}-2k-1|>\frac{1}{4(\sqrt{m}+1)}\Leftrightarrow \frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}.
\]](http://latex.artofproblemsolving.com/1/e/0/1e09c9abb86cd903b051995f222370c8fd8de90e.png)
Note that 
,![\[
\frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}\ge \frac{2}{2\sqrt{m}+2k+1}\ge \frac{2}{4\sqrt{m}+1}>\frac{1}{4\sqrt{m}+4}
\]](http://latex.artofproblemsolving.com/6/8/1/68110c4cbf0e70a166d6d3e86b0d0b2b67d905d8.png)
for trivial reasons.
.
,
.![\[
\frac{1}{2\sqrt{k^2+k}+2k+1}>\frac{1}{4\sqrt{k^2+k}+1}\Leftrightarrow 2\sqrt{k^2+k}>2k,
\]](http://latex.artofproblemsolving.com/5/4/0/54096d53aec6ebbfd77f9412c2e38a786183f4d0.png)
which is clearly the case.
,![\[ \sqrt{m} - \frac{1}{2} \ge \frac{1}{2} > \frac{1}{16} \ge \frac{1}{8(\sqrt{m} + 1)} \]](http://latex.artofproblemsolving.com/6/c/8/6c85769fed2b1b165b05647f2d02b43cb052da0e.png)
