[/quote]
,
have the property that both![\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]](http://latex.artofproblemsolving.com/d/e/c/dec527c4673470e7023d6aed00fac90268bd2fcc.png)
are consecutive perfect squares.
be real numbers such that 
and 
Prove that
Where 
be real numbers satisfying![\[
x_1 + x_2 + \cdots + x_6 = 6,
\]](http://latex.artofproblemsolving.com/6/2/2/622a329112ce9613a5c389b4c9078cd43c1c3f05.png)
![\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]](http://latex.artofproblemsolving.com/8/f/a/8fa3f74ba7c3587576116a7bdbb4e9f88d389133.png)
Find the maximum possible value of the product![\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]](http://latex.artofproblemsolving.com/3/b/4/3b48315e2775b4ac3913d7c9b02dbc81daed8d15.png)
be an acute triangle. Let 
be the reflection of point 
over the line 
.
and 
be the circumcenter and the orthocenter of triangle 
be the midpoint of segment 
.
and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle 
is a point on the line 
such that 
,
.
be real numbers such that 
. Find the value of 
and find at least one triplet 
that satisfy those conditions.
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
such that there exists a sequence of positive real numbers 
,
,![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
of the set 
,
such that:
,
,
.
such that:
is strictly increasing,
such that 
for all 
,
,
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
be a prime number and 
.![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
,
,
,
of real numbers satisfies![\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]](http://latex.artofproblemsolving.com/4/d/a/4da3b88a20c42c1798141b8db086de341cfb9a67.png)
for every positive integer 
such that 
for every positive integer 
such that 
is divisible by 
and 
.
Y by MOM-Admin, applepi2000, Chandrachur, JustN, socrates, WJ.JamshiD, rkm0959, don2001, anantmudgal09, Davi-8191, tenplusten, adityaguharoy, Mathuzb, Tawan, itslumi, centslordm, HWenslawski, megarnie, Mathlover_1, ImSh95, tiendung2006, Adventure10, Mango247, Deadline
Let 
be the set of real numbers. Determine all functions 
that satisfy the equation![\[f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\]](//latex.artofproblemsolving.com/6/b/2/6b2acc0a1d76866e461fdf5d96126806a9ead8dd.png)
for all real numbers 
and 
.
Proposed by Dorlir Ahmeti, Albania
be the set of real numbers. Determine all functions 
that satisfy the equation![\[f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\]](http://latex.artofproblemsolving.com/6/b/2/6b2acc0a1d76866e461fdf5d96126806a9ead8dd.png)
for all real numbers 
and 
.Proposed by Dorlir Ahmeti, Albania
This post has been edited 5 times. Last edited by djmathman, Jun 14, 2018, 4:22 PM
Reason: italicized author
Reason: italicized author
,
,
,
in (1), we can get
,
.
is a solution.
.
be the set of all fixed points.
from 
and 
from 
with 
,
,
or 
.
,
.
,
with 
,
From (3) and (5), we know that 
;
,
.
,
.
for 
,
,
,
or 
tha is true but the moment you said
so 
this is wrong that's why
this mean 
but than
we have 
so 
this mean
or 
if you prove 
be the assertion 
or 
,
.
for all 
.
when 
;
when 
.
be some zero of 
,
.
.
: 
.
.
: 
(by setting 
)
.
and 
(follows from 
)
.
.
and apply 
to get at the end 
,
.
that 
,
.
(case 1) and 
(case 2).
![\[ (1) \quad f(f(y))+f(0)=f(y)+yf(0), \]](http://latex.artofproblemsolving.com/d/e/9/de97b729878e146f86328118521853fa176e6c94.png)
![\[ (2) \quad f(x+f(x))+f(0)=x+f(x), \quad \quad f(x+f(x+1))=x+f(x+1) \]](http://latex.artofproblemsolving.com/7/c/8/7c85287c125204e947588822558fe63ec883832c.png)
and there exists fixed point
.
.
,
or 
.
,
we have 
.
.
.
,
.
and 
are two functions satisfying the conditions of problem.
?
gives 
fixed point and we still have 
fixed point, and so 
is 
so if we can somehow show that 
and we'd be done.
into 
and somehow messed it up.
gives 
.
and 
gives 
.
and 
.
if and only if 
.
gives 
,
.
.
.
gives 
and 
gives
This means that 
.
gives
so 
Therefore, if 
,
,
.
gives 
and 
gives 
,
gives 
,
,
.
,
,
.
.
,
,
,
.
gives 
,
.
.
and 
.
and 
implies 
.
implies 
.
implies 
and so 
implies 
.
implies 
,
and 
implies 
,
.
and that 
which implies that 
hence 
.
.
for all 
yields 
.
gives that 
and since 
,
.
yields 
.
implies that 
,
here gives 
,
and 
,
gives 
.
for all integers 
,
,
,
,
.
,
and 
for all integers 
gives 
,
gives 
,
.
,
as well. However since for 
,
is in 
,
implies that 
.
and hence 
,
for all 
and 
we have that 
,
for all integers 
for any 
.
gives 
,
.
implies that 
,
as well.
gives 
.
,
.
for all 
,
,
for all integers 
gives 
,
or 
We can check these easily works. Now we prove they are the only ones.
then, 
or 
Thus we have two cases.
gives
we let 
in 
,
gives,
then 
.
then
such that 
The base case for positive integers is 
,
.
in 
,
Then, fix 
we get 
Hence, 
Completing the proof..
be a fixed point if and only if 
fixed point
fixed point
The problem is reduced to proving that 
,
is a fixed point.
are fixed points then so is 
.
fixed poin
is a fixed point then 
we find that 
.
we find that 
we find that 
,
we obtain 
be the set of fixed points, i.e. 
.
.
,
.
for all 
,
Then by 
,
for all 
so we're done. 
as our only solution
,
and 
:![\[
(x,y) \rightarrow (0,x) \Rightarrow f(f(x)) + 2 = f(x) + 2x \quad (1)
\]](http://latex.artofproblemsolving.com/a/8/2/a82c5c4ada361b2f23326b2062f67afc3627e86a.png)
![\[
(x,y) \rightarrow (x-1,1) \Rightarrow f(x + f(x) - 1) = x + f(x) - 1 \quad (2)
\]](http://latex.artofproblemsolving.com/e/2/f/e2f666541d620e3093b74a16908e89bd88d49381.png)
,
gives 
,
.![\[
(x,y) \rightarrow (0,x) \Rightarrow f(f(x)) = f(x) \quad (3)
\]](http://latex.artofproblemsolving.com/b/d/c/bdcffe2c194dd50136a2af7ef9a47a1cdfa7babd.png)
![\[
(x,y) \rightarrow (x,0) \Rightarrow f(x + f(x)) = x + f(x)
\]](http://latex.artofproblemsolving.com/3/8/8/3885d360a7e00f89ece895417fedc74e6881e15e.png)
![\[
(x,y) \rightarrow (x,-x) \Rightarrow f(x) + f(-x^2) = x - xf(x) \quad (4)
\]](http://latex.artofproblemsolving.com/9/b/5/9b56112353424a2bb1bdd376c8ab5469b02bc0dd.png)
![\[
(x,y) \rightarrow (-x,x) \Rightarrow f(-x) + f(-x^2) = -x + xf(-x) \quad (5)
\]](http://latex.artofproblemsolving.com/7/3/8/738258d4a7df4668589634b91761af1eca0c88f1.png)
,![\[
f(x) - f(-x) = 2x - x(f(x) + f(-x)) \quad (6)
\]](http://latex.artofproblemsolving.com/0/6/6/066984ad7fad66ad96c9be1d7567e17bd4b06037.png)
![\[
(x,y) \rightarrow (x,f(x)-x) \Rightarrow f(x(f(x)-x)) = (f(x)-x)(f(x)) \quad (7)
\]](http://latex.artofproblemsolving.com/9/1/9/919fad13b3686a8f67b44ea3127f97a3926789f4.png)
![\[
(x,y) \rightarrow (-x,x-f(x)) \Rightarrow f(-x(x-f(x))) = (x-f(x))(f(-x)) \quad (8)
\]](http://latex.artofproblemsolving.com/4/8/a/48a96411ca9fe3baa0f09388b170256eb745a3dc.png)
in conjunction with 
,![\[
-f(x)(f(x)-x) = (f(-x))(f(x)-x)
\]](http://latex.artofproblemsolving.com/7/5/1/7512a0cc5cae401d71b4817c070925fbc6fd8641.png)
or 
.
.
.
.
.
.
.
.
.
.
.
.
.
.
are the only solutions, both of which can be plugged in and checked to work easily.
.![\[f^2 (y) + f(0) = f(y)+ yf(0).\]](http://latex.artofproblemsolving.com/4/0/9/4092c3b7c1c910a86abb5718df5cf72b9d5cb3cd.png)
In particular, taking 
.
,
or not.
gives us 
.
gives us
where we use our results from item 1 to simplify 
as 
,
gives us![\[f(f(C)) = f(C) + 2C - 2,\]](http://latex.artofproblemsolving.com/a/f/4/af44d3b3f553f63f35a7a6187a6911bc05e9824f.png)
and solving for 
as 
.![\[ f(x+f(x+1))= x+f(x+1),\]](http://latex.artofproblemsolving.com/a/1/6/a1698d18147344c9d6edf9088b7a866aee5a0225.png)
so 
is a fixed point, implying
and 
.![\[f(x) + f(-x^2) = x - xf(x).\]](http://latex.artofproblemsolving.com/2/5/5/255d1cd647673484e963e879a1f8234ffe97b70d.png)
Setting 
are fixed points, then 
is a fixed point.
gives us
![\[ f(x-1 + f(x)) = x-1 + f(x),\]](http://latex.artofproblemsolving.com/e/1/4/e14ac13395ce88467933120e00cd9ca6bf43e8f0.png)
so 
is a fixed point of 
gives us![\[ f(x+f(x)) = x + f(x),\]](http://latex.artofproblemsolving.com/5/9/c/59c4333891de41f121ce74a723601230d187b7d7.png)
so 
gives us![\[f(x+f(x)+1) + f(-x-1) = x+1 + f(x) - f(x+1).\]](http://latex.artofproblemsolving.com/f/5/c/f5cbb7e0bd644d3779864c8ad87164e63b570010.png)
Since 
for all 
and ![\[\begin{cases}f(x) + f(-x^2) &= x - xf(x) \\ f(-x) + f(-x^2) &= -x + xf(-x).\end{cases}\]](http://latex.artofproblemsolving.com/5/3/2/532fbf19b3744db1a933fee359b2e672d1d06183.png)
Subtracting these two equations gives us ![\[2f(x) = 2x \implies f(x) = x,\]](http://latex.artofproblemsolving.com/7/d/9/7d9d0f297b077dd890782d91b5ced40085c9f01a.png)
so we are done.
First let's suppose 
So, 
putting 
in 
Comparing this and 
But also 
So from 
we get 
for all 
so it follows that 
for all 
in 
we get 
,
note that this implies 
Let 
.
gives us 
,
.
.
implies that 
,
.
so we have in any fixed point 
.
in terms of 
.
and we by this
and if we plug it in 
.
and use the value of 
and 
that get compute them already you can find:
.
for regarding of these two case we have for all real numbers 
