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)
.
and ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a strictly monotonous function such that 
.
.
for which there exists a scrictly monotonous function 
such that 
,
, ∀x∈[0,1].
Y by shinichiman, anantmudgal09, Adventure10
A positive integer 
is called a Jane's integer if 
for some positive integers 
and 
. Prove
that for every integer
there exist infinitely many positive integers 
such that the set of 
consecutive
integers
contains exactly 
Jane's integers.
is called a Jane's integer if 
for some positive integers 
and 
. Provethat for every integer
there exist infinitely many positive integers 
such that the set of 
consecutiveintegers
contains exactly 
Jane's integers.
contains at most 
Jane's integers.
with 
in 
then 
.
then from 
we find 
or ![$4y^2>i \left[ 3(2t-i)^2+i^2 \right] \ge 3it^2$](http://latex.artofproblemsolving.com/0/1/e/01e22ba3b89cda14a388bc378ca2aacdb38ad551.png)
or 
.
Therefore, there are at most 
Jane's integers 
in 
.
then since 
so there are at most 
Jane's integers 
in 
(The proof for the bound 
can be found 
and there are at most 
Jane's integers in 
so that their difference is at least 
.
and we are done. 
Jane's integers.
then we can see that the set 
already has 
.
is also a Jane's integer, or 
for some 
.
.
then 
.
so 
and we have 
.
with fundamental solution 
and has infinitely many solutions 
:
From this, we find that there exists infinitely many solutions 
for Pell's equation so 
,
so 
be number of Jane's integers in set 
.
so 
and 
then there must exists 
so 
.
.
,![$[1,N^6]$](http://latex.artofproblemsolving.com/7/6/3/76304f103518b4deb9ee3987b601176e01d31d34.png)
is at most 
where 
is at most 
but also at least 
.
for all positive integer 
and fixed positive constant 
.
,
let 
denote the number of good numbers in the set 
.
so 
is "discretely continuous", thus it suffices to show that both 
and 
hold infinitely often to get the desired result.
for some integer 
.
for 
is good. Also, let 
satisfy the Pell-ish equation, 
.
so we get another good number, making 
is possible infinitely often, take the interval ![$[1, x]$](http://latex.artofproblemsolving.com/8/2/c/82c624937a75943fff85c58d5c06ec7d843e67f4.png)
and observe that ![$$a^3+b^2=c \leqslant x \Longrightarrow (a, b) \in [1, x^{\frac{1}{3}}] \times [1, x^{\frac{1}{2}}],$$](http://latex.artofproblemsolving.com/9/0/f/90f872a1e4007fda8026c4d95921fae4161fc8be.png)
so the density of good numbers is at best 
However, if 
for all but finitely many 
contradiction!
and any 
?
,
,
and the Pell argument is essential there.
