[/quote]
,
be non-negative real numbers. Prove that![\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)^2}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \left(\sum_{i=1}^n a_i\right)^2 - \sum_{i=1}^n a_i^2.\]](http://latex.artofproblemsolving.com/b/1/6/b16f7db4a2f1bb30df4f6534bc6fa18b0fd6b01b.png)
Assume all denominators are non-zero.
be non-negative real numbers. Prove that![\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \sum_{i=1}^n a_i.\]](http://latex.artofproblemsolving.com/5/c/2/5c2e78940e212cd290f76cc566b8c8018c8fea44.png)
Assume all denominators are non-zero.
,
for all positive 
and 
and 
be two triangles with a common orthocenter 
.
are concurrent
Prove that
Prove that
,
,
,
be reals, and let 
.![\[ f(5,5)f(5,6) = f(2,4)f(3,4) = f(1,4)f(4,4) \implies f(1,1)f(1,2)f(4,5)f(4,6) = f(2,3)f(3,3)f(1,5)f(1,6) \]](http://latex.artofproblemsolving.com/b/9/6/b9663a84c82661cb6a5e1cfdf3894c9943892a6f.png)
be an acute triangle and let 
be the midpoint of 
.
passing through 
and 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that 
.
of positive integers for which the numbers 
and 
are both positive integers.
![\[\tau\big(\varphi(n)\big) \geq \varphi\big(\tau(n)\big)\]](http://latex.artofproblemsolving.com/1/e/f/1ef41ab06470c85fd23226c2b470b1811629b559.png)
with all integer n.
of integers is defined by![\[
-\frac{1}{2} < a_{n+1} - \frac{a_n^2}{a_{n-1}} \leq \frac{1}{2}
\]](http://latex.artofproblemsolving.com/5/2/f/52ffcd88e92af6bbff0073c8f1a536cfd4644a9a.png)
with 
,
,
is odd for all values of 
.
grid of unit squares. Two distinct unit squares are adjacent if they share a common side. Each unit square is to be coloured either black or white. Such a colouring is called 
if every unit square in the grid is adjacent to an even number of black unit squares. Determine the number of 
be positive integer such that ![\[ \text{LCM}(x+2, y+2)+\text{LCM}(x, y)=2\text{LCM}(x+1, y+1).\]](http://latex.artofproblemsolving.com/7/3/3/7338e4d6058ededa9b8cd9fcc945ae362db4bad4.png)
Prove that 
Y by GeoKing
A finite set 
of positive integers is called cardinal if contains the integer 
where denotes the number of distinct elements in . Let 
be a function from the set of positive integers to itself such that for any cardinal set , the set 
is also cardinal. Here denotes the set of all integers that can be expressed as 
where 
. Find all possible values of 
Proposed by Sutanay Bhattacharya
of positive integers is called cardinal if contains the integer 
where denotes the number of distinct elements in . Let 
be a function from the set of positive integers to itself such that for any cardinal set , the set 
is also cardinal. Here denotes the set of all integers that can be expressed as 
where 
. Find all possible values of 
Proposed by Sutanay Bhattacharya
This post has been edited 2 times. Last edited by math_comb01, Jan 21, 2024, 12:48 PM
is trivially a solution. Assume non-constant functions 
is the only such solution. We induct on 
.
which forces 
.
.
.
is also cardinal, which forces 
.
for all 
and 
for 
for some 
work?
and 
?
which can be proved by lots of casework.
gives a cardinal set every time. This is rather easy to note. Let's take 
.
.
.
where 
.
or 
but not both. As 
can be any real number, this is claiming that 
which will be false the moment we consider 
.
as a solution.![$f(x)=1, x \in [1,2023] \cup [2025, \infty]$](http://latex.artofproblemsolving.com/8/b/9/8b95950174cb6b3f96099c460f23c43dbee44070.png)
and 
also works. We cannot repeat this for more numbers as the set 
will get violated.
and 
?
?
.
is 
,
is 
and 
is 
.
.
,
be the smallest positive integer with 
.
.
.
cardinal, so 
(because if 
it would have 
,
.
is also cardinal. This implies that 
for each 
.
,
is cardinal. Since 
are fixed points, we have 
is cardinal. If 
,
,
is cardinal, which is false since it has 
elements. Therefore, we have 
and 
.
,
,
and get 
is cardinal, which is absurd. If 
,
gives that 
is cardinal, which is false. If 
,
,
gives that 
must be cardinal, which is false.
,
and 
for all 
,
for all 
,
.
,
,
,
,
must either be 
,
must either be 
.
.
.
.
,
.
,
,
,
.
.
,
.
,
.
,
a contradiction. If 
,
,
is required.
, but probably yes if you aren't in 12th. If you are in 12th, I don't know.
:
if and only if 
.
.
classes, we can choose a set 
of 
.
be part of an infinite equivalence class. Choose a set 
,
.
integers satisfying 
is cardinal, and its image under 
.
.
is cardinal 
(other than 2), we can say that 
is also cardinal forcing 
,
.
or 
is cardinal 
is also cardinal forcing 
or 
,
.
.
,
.
.
is cardinal. But that means the set 
is cardinal which is clearly false (only 
distinct elements). Contradiction, so 
.
is cardinal, so 
is cardinal.
is cardinal. This set has 
.
.
and 
or 
,
.
,
where 
and 
.
is cardinal. So, 
,
is not cardinal, a contradiction. 
.
.
.
,
is not cardinal, a contradiction. 
by lemma 1, and 
either 
.
,
where 
.
,
and 
.
,
and 
.
and 
for all 
.
.
and the latter cannot be a cardinal set, a contradiction. 
,
to be consistent with lemma 2.
,
but the latter is not cardinal, which is impossible.
,
but the latter is not cardinal, which is impossible.
,
but the latter is not cardinal, which is impossible.
as another possibility.
,
or 2 ,
or 2.
as if 
,
is not cardinal.
using stong induction on 
being trivial. ASSUME holds for 
.
,
.
is cardinal. If 
So, set 
is not cardinal, though it was to be cardinal as 
is cardinal. Thus 
in this case.
.
,
,
is clearly not cardinal.
,
,
,
is clearly not cardinal.
,
is clearly not cardinal.
as
and 
.
or 
.
,
. Clearly these constructions are valid, now let's prove 
,
means that 
is a set which contains every distinct element of a multiset 
.
that satisfy
. Since 
is cardinal, the set 
is also cardinal. Clearly 
where there are infinitely many 
that satisfy 
.
that satisfy
. Note that 
is cardinal. If 
,
is also cardinal which forces 
,
,
yields 
.
,
such that
. Note that the set 
is cardinal. If 
for some 
is cardinal. A contradiction, that means 
for all 
.
is cardinal, 
.
is cardinal, so 
.
is cardinal, so 
.
.
.
is cardinal, so 
.
is cardinal, so 
.
,
are cardinal, 
and 
or 
is as well, which implies that 
.
since then we can take 
which would mean that 
is cardinal which is contradiction. So 
.
as "ascending".
is ascending until a term that is less than or equal to the previous term.
,
,![$1, 2, 3, \dots, k, 1, [\text{some sequence of 1's and 2's}], x$](http://latex.artofproblemsolving.com/2/4/9/24942d4baf3dbf49d4d943b318cf8564eef2d256.png)
for some 
.
makes a contradiction. qed
to get 
.
.
.
gives contradiction for 
,
gives contradiction. For 
,
.
.
has 
elements, so 
has at most ![\[
1\;\le\;f(n)\;\le\;n.
\]](http://latex.artofproblemsolving.com/5/b/c/5bc37bcc9dab0611663c49b1ffbdc61909f00c7e.png)
is injective, then 
(apply to 
)![\[
f(n)=n
\quad\forall n,
\]](http://latex.artofproblemsolving.com/4/a/f/4af5e1e65ffd483160b683c1836c0651aa91ac0f.png)
and in particular 
.
with![\[
f(a)=f(b)=r>1.
\]](http://latex.artofproblemsolving.com/3/1/0/310a283641ed97eaf39534560bf1780a0ec14537.png)
Consider the set![\[
T=\{a,a+1,\dots,b\},
\]](http://latex.artofproblemsolving.com/9/a/e/9ae6b18c7824ef9f56ed65dc8971b034ebdc80ae.png)
which has 
.
is at 
and 
,![\[
\bigl|f(T)\bigr| \;=\; |T|-1 \;=\; b-a.
\]](http://latex.artofproblemsolving.com/6/3/2/6322b16e69a0fa1df248e842a9c231d682c2a707.png)
But 
must be cardinal, so 
.
,![\[
b-a \;=\; r.
\]](http://latex.artofproblemsolving.com/2/b/7/2b7f5cdbd35d0262e729d023316dc675b39957a5.png)
Hence any repeated value must equal the distance between its two preimages. In particular, if 
then 
but one checks this clashes with the bound 
.![\[
r=1\quad\text{or}\quad r=2,
\]](http://latex.artofproblemsolving.com/4/2/0/4206c4e53486e58a489c578f0d405943d18d9ba9.png)
and so![\[
f(n)\in\{1,2\}
\quad\text{for all }n.
\]](http://latex.artofproblemsolving.com/a/d/b/adbbf9f686439e875f92c80b1838e74d21a6f76d.png)
In particular 
.![\[
\boxed{f(2024)\in\{1,2,2024\}}.
\]](http://latex.artofproblemsolving.com/5/0/e/50e5a7cfb945942dad3b5be52cf01b7d45ef71af.png)
