to satisfy that the equation 
has at least 1 roots 
satisfying the following condition: for every monic polynomial 
of degree at most 
and 
distinct integers 
such that ![\[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\]](http://latex.artofproblemsolving.com/e/e/8/ee8adc79dc258b021529245fec16e081da2d1b0a.png)
.
and for 
,
Prove that every natural number occurs exactly once in this sequence.
inscribed in 
,
is the midpoint of arc 
of 
intersects 
at 
.
intersects 
at 
different from 
.
is the orthocenter of triangle 
= 
Y by Rohit-2006
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form
)
![\[ f(f(x)^2 + f(y)) = x f(x) + y \]](http://latex.artofproblemsolving.com/d/0/2/d022cecec0757daff00434c6a686225ccf384eef.png)
.
to get 
for 
.
for 
and 
.
to yield
We then have
If 
and 
for nonzero 
and 
,
This makes 
or 
,
to find 
.
is not constant, we have 
and 
.
denote the assertion, and observe that there are no constant solutions. By 
,
,
,
.
,
is nonconstant, we must have 
.
,
,
,
,
,
,
for some real 
is the only solution.
![\[ f(0) = 1 \]](http://latex.artofproblemsolving.com/8/7/8/8784a89d399b9480a41862bb77e7645094282249.png)
![\[ f^k(x) = f(x\sqrt{k}) \quad \forall k \in \mathbb{N}, \forall x \geq 0. \]](http://latex.artofproblemsolving.com/7/3/1/731d31076d43a297e6a9b249cfffd88bf5871f18.png)
,
and 
if 
gives injective and varying 
.
is a constant value, and since 
and therefore 
.
.
gives that 
,
implies that 
,
.
for some constant 
,
.
from 
to reals where 
and $f^k(x)=f(x\sqrt k).
and 
and 
,
Since 
and 
,
This makes 
.
.
with 
in the given functional equation, we get![\[
g(x)^k = g(kx).
\]](http://latex.artofproblemsolving.com/e/c/2/ec229f3427eadce49a2b42957cd6cf6ee4210516.png)
also we can prove that it's positive always by contradiction, assume 
a 
,
> 0 take a sequence 
as n tends to 
from this we can contradict continuity therefore 
![\[
h(x) = \log g(x).
\]](http://latex.artofproblemsolving.com/7/e/f/7ef39ee05c2bf39ad71ee6a16459195daa8cf2b6.png)
,
.
and
,
,
that forms an Arithmetic Progression
Now, set 
Subtracting the two equations, we get that:-
for some 
value and 
be the set of real numbers. Determine all functions 
such that![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
such that 
and![\[
\Bigl(f(ab)\Bigr)^2 = \Bigl(af(b)+f(a)\Bigr)\Bigl(bf(a)+f(b)\Bigr) \quad \forall a,b \in (0,1).
\]](http://latex.artofproblemsolving.com/0/f/2/0f2d0a333bdd73ff468072cfd716f94865385710.png)
of pretty well known FEs 
we see 
.
and since 
for arbitrary 
we see 
.
and since 
is the identity.
works for all real 
.
twice to get 
which is equivalent to 
gives 
for nonnegative real 
gives 
(Cauchy FE) for nonnegative real 
so it still works for all nonnegative real 
for all nonnegative rational 
so 
be a positive rational
so 
which is equivalent to 
so 
.
such that![\[ f(x^2 f(y)^2) = f(x)^2 f(y) \]](http://latex.artofproblemsolving.com/d/9/3/d93711733d0b34679f88e7a4d5fa00470dd539d4.png)
for all 
,
denote the assertion.
,
,
.
;
.
,
.
.
is the square of a rational number. Suppose 
-
and 
are 
-
for every 
.
