AoPS Academy
be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
Find the minimum value of 
be real numbers such that
Prove
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South Africa
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.
). And indeed, the 
should be 
; mind you, when it works for the latter, it will a fortiori work for the former! of the quadrilateral 
is 
, by the known fact that the area of a convex quadrilateral is half the product of its diagonals with the sine of their angle, followed by the fact that the sine of an angle is at most , followed by Ptolemy's inequality.
, so if we prove that 
, i.e. 
, that will be enough. But this is a simple vectorial computation.
and 
, so 
.
, by the triangle inequality.
. Then 
, 
, 
, so the inequality is equivalent to 
, which, as far as I remember, is the triangle inequality, expressed in the Euclidean norm (in any dimension) by Minkovski's inequality. The fact it is not appearing in its usual form
,
(having as last recourse the triangle inequality - what else?); otherwise it is difficult to foresee a change in the chain of formulae.
![$\max_{a,\,b,\,c>0\atop a+b+c=1}(a+\sqrt{b}+\sqrt[3]{c})$](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 
,
be a convex quadrilateral and points 
and 
on sides 
such that![\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]](http://latex.artofproblemsolving.com/b/4/2/b42cd6305fab848a60c79689f17bdb55e1ba0d31.png)
![\[\tfrac{AB}{AE}=\tfrac{CD}{CF}=n\]](http://latex.artofproblemsolving.com/f/8/2/f82128eb770639038420fce432811dd37da6243d.png)
and 
than we have 
hence a mistake in question?
instead of 
it seems.
such as 
are parallelograms.
collinear 
such as 
.
.
.
.
and 
be perpendicular lines on 
.
,
so 
.
inequality is equivalent with 
.
.
and 
.
.