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such that:![\[ f(xy + f(x))+f(y) = xf(y)+f(x+y) \]](http://latex.artofproblemsolving.com/6/0/4/604d965a38e7df743b97369340946ef8d1c56e5e.png)
for all real numbers 
such that for every integer 
with 
is satisfied that :
is odd
be a positive integer. Find the minimum possible value of![\[
S = 2^0 x_0^2 + 2^1 x_1^2 + \dots + 2^n x_n^2,
\]](http://latex.artofproblemsolving.com/7/b/8/7b818fccfea8f1dcb9f80bde51dba224f91c3713.png)
where 
are nonnegative integers such that 
.
, let 
. For each subset 
of 
, assign a non-negative real number 
, and define 
, and for any subsets 
, whenever 
, 
Determine the minimum possible value of![\[\sum_{I \subseteq S} (-1)^{n-|I|} Y_I^{k},\]](http://latex.artofproblemsolving.com/6/e/f/6ef80d82ee22435b9b2994ec60790860621d9aea.png)
where 
denotes the number of elements in the finite set .
be the unit group of 
. It is well known that this group is cyclic. Let 
be a generator of this group and consider the map 
for a fixed positive integer . I know that the kernel 
has oder 
. By the first isomorphism theorem, 
. Since 
is cyclic, so are its subgroups and hence 
is cyclic of oder 
. Let 
. Then 
is cyclic too and hence we have the partition:
(for example ).
have non-trivial solution for all primes 
. Here is my attempt:
, as 
(the case 
is easy).
, then everything is a cube (since the cubing map 
is an anutomorphism by above) and the equation is solvable.
, let 
and 
be the cosets where 
, then we have the following cases:
or 
, Without loss of generality, assume 
is a cube, then 
or 
and we may take 
as our solution.
and 
, then 
and 
and we may take 
as our solution.
and 
, then 
and 
and we may take as our solution. and , then and . This is the case I am stuck with. If we have 
, then we may take 
as our solution since 
for and 
is not 
). But it is not always possible to have both and . For example, I can take 
, then , but 
iff 
.
.
is a normal matrix, then 
is unitarily similar to 
,
is hermitian and 
is skew-hermitian. necessarily normal?
is an ordered field with all field axioms holding.
Suppose 
is a real number and 
(Set of real numbers). Prove that if 
. Then it (that is 
) must be .
is a real number. If 
be any complex number and 
. Show that if 
, then 
.
Order property is not true for all complex numbers.
, for 
, 
with 
, 
.
?, 
is the length of 
, hence 
, define 
as the smallest positive integer satisfying the following property: for any integer 
coprime with , we have 
.
, and integers 
all coprime with , prove that there exists a non-empty subset of 
such that![\[\prod_{i \in I} a_i \equiv 1 \pmod{n}.\]](http://latex.artofproblemsolving.com/8/f/a/8fadde1e14e28dda0cb2371b69e783fd7e572352.png)
Proposed by Zhenqian Peng from High School Affiliated to Renmin University of China
and 
functions are given with following properties:
is strict increasing and for each 
there holds 
or 
.
if holds and 
if holds.
and 
such that for all 
we have 
![$f:[-1,1]\to\mathbb{R}$](http://latex.artofproblemsolving.com/5/a/e/5aee4d7ba286e59465514a88e4d50c68e96d2a4b.png)
be a continuous function such that
for every 
in ![$[-1,1],$](http://latex.artofproblemsolving.com/3/a/9/3a9a9276194df4e3ca7e915b63a202e7920fd4a4.png)
and
exists and is finite. is unique, and express 
in closed form.
be an integer, 
a real number, and 
be positive real numbers such that 
. Prove that:![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
:
we have:
again for entire 
:
![$\max_{a,\,b,\,c>0\atop a+b+c=1}(a+\sqrt{b}+\sqrt[3]{c})$](http://latex.artofproblemsolving.com/e/d/8/ed81d27343068e602406c53746b21abdb401f44a.png)
Suppose 
then multiplying 
iff 
iff 
iff 
iff 
this contradicts order axiom of real numbers. A similar argument can show that if 
we again get a contradiction to order property and thus since 
is the only valid choice.
(Since 
iff 
,
.![\[
\sum_{{\rm cyc}}\frac{1+x_i^k}{1+x_{i+1}}\ge 2^{1-k}\sum_{{\rm cyc}}\frac{(1+x_i)^k}{1+x_{i+1}}.
\]](http://latex.artofproblemsolving.com/e/c/c/eccce7f89c7b8c2c46ce0bf15019c9451199ef36.png)
Next, by AM-GM inequality![\[
\sum_{{\rm cyc}}\frac{(1+x_i)^k}{1+x_{i+1}}\ge n\sqrt[n]{\prod_{1\le i\le n}(1+x_i)^{k-1}}\ge n\sqrt[n]{\prod_{1\le i\le n}2^{k-1}x_i^{\frac{k-1}{2}}} = n\cdot 2^{k-1}
\]](http://latex.artofproblemsolving.com/8/6/f/86f982a637e94b51f1c73ca79bb0b3511960beeb.png)
using 
.
![$\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n \sqrt[n]{ \frac{(1+x_1^{k-1})...(1+x_n^{k-1})}{2^n}} \geq n $](http://latex.artofproblemsolving.com/c/2/0/c20cf80bed2989e2ec4cc2ec3752b83e7239fec5.png)
,
,
which, together with AM-GM Inequality as well as Hölder’s Inequality, implies that![\begin{align*} & \frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \\
\ge&n\sqrt[n]{ \frac{\left(1 + x_1^k\right) \left(1 + x_2^k\right)\cdots \left(1 + x_n^k\right)}{(1 + x_1)(1 + x_2) \cdots (1 + x_n)}}\\
\ge&n\sqrt[n]{ \frac{\left[\left(1 + x_1\right)\left(1 + x_2\right)\cdots \left(1 + x_n\right)\right]^{k-1}}{2^{n(k-1)}}}\\ \ge&n\sqrt[n]{ \frac{\left[\left(1 + \sqrt[n]{x_1x_2\cdots x_n}\right)^n\right]^{k-1}}{2^{n(k-1)}}}=n,
\end{align*}](http://latex.artofproblemsolving.com/5/e/6/5e6fbf0dd971176f090ff757b4d54093c4bfc23c.png)
and the minimum value 
.