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.
such that:
be a positive integer, and let 
denote the integers modulo . Determine the number of functions 
satisfying 
for all 
.
such that 
divides 
for any integers 
and there exists a polynomial 
with integer coefficients, such that 
for all 
.
be defined by 
(a) Show that 
has exactly one point of discontinuity. at its point of discontinuity.
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
for which: Every sequence, there always exist two terms whose diffence is not less than 
. (where is given positive integer)
![$f \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/7/a/a/7aa8f551ba6422af33ab978aaf842329b6c2aced.png)
such that its Fourier series diverges at 
. There is nothing special about the point , it was just for convenience. The same proof showed that for every ![$t \in [-\pi,\pi]$](http://latex.artofproblemsolving.com/d/1/3/d1331fab5626e64ec75c6c40edc6379bc591c803.png)
, there is ![$f_t \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/e/1/e/e1ef8ac797a9743ebd9e71bf1682b0246b221086.png)
such that the Fourier series of 
diverges at 
, not to show.
be a Banach space and for every 
we have a normed space 
. Suppose that for every there is 
and 
such that![\[
\sup_{k \geq 1} \| T_{n,k}x_n \|_{Y_n} = \infty.
\]](http://latex.artofproblemsolving.com/a/1/b/a1b7a2453200f95f61c2451fc91a8fc3a81fd8fd.png)
Show that![\[
B = \left\{ x \in X : \sup_{k \geq 1} \| T_{n,k}x \|_{Y_n} = \infty \ \forall n \geq 1 \right\}
\]](http://latex.artofproblemsolving.com/4/1/a/41a671b6c40492f454bf79d1a3205cd5592a94e6.png)
is of second category. (I am given the hint to write 
as![\[
A = \bigcup_{n \geq 1} A_n = \bigcup_{n \geq 1} \left\{ x \in X : \sup_{k \geq 1} \| T_{n,k}x \|_{Y_n} < \infty \right\}
\]](http://latex.artofproblemsolving.com/b/9/1/b9174ec68dd2a503c6d68deaf281c5e110637cbd.png)
and show that 
is of first category.)
. Show that there is ![$f_D \in C_{\text{per}}([-\pi,\pi]; \mathbb{C})$](http://latex.artofproblemsolving.com/8/a/b/8ab83356343e4c5e30778770377027d0b2d57735.png)
such that the Fourier series of 
diverges at 
for all . (I'm given the hint to use part a) with![\[
T_{n,k} : (C_{\text{per}}([-\pi,\pi]; \mathbb{C}), \| \cdot \|_\infty) \to \mathbb{C}, \quad f \mapsto S_k(f)(t_n),
\]](http://latex.artofproblemsolving.com/2/2/5/225a5eec7a9582981841918fc28de96379b31404.png)
where![\[
S_k(f)(x) = \sum_{|j| \leq k} c_j(f) e^{ijx}.
\]](http://latex.artofproblemsolving.com/a/c/3/ac37ad818424867f4504e1472b2abe0883b6284b.png)
has circumcircle 
and incircle 
, where is tangent to 
at 
, respectively. 
is an arbitrary point on . 
meets 
at 
, 
meets again at , 
meets again at 
. 
is a point on such that 
are colinear. Let 
be the intersection of 
and , prove that 
is tangent to .
. This set appears to be empty for 
, so there are no 0-national matrices, let alone invertible ones.
now. The matrices to be counted are of the form
where the inequality is equivalent to the matrix being invertible. The inequality is equivalent to 
. If we ignore the inequality for a moment, we have 
possible matrices. We must subtract those with 
. Here it becomes obvious that it does not matter if 
are in the range from 
to for a positive or in the range from 
to for a negative . Let therefore be positive., the possible sums range from 
to 
. For the sum 2, there is one choice for 
to produce that sum. For 3, there are two choices. We reach the maximum number of choices for the sum 
where we have choices. From then onward, we incrementally go down to one choice for the sum .
and 
, we must square each number of choices to obtain the number of combined choices for both sides of 
. Therefore, the total number of choices with 
is
Therefore, for a positive , the number of invertible n-national matrices is
For a negative , the variable needs to be replaced by 
here.
such that 
then find minimal value of 
.
and their entries belong to the set 
For all 