and 
.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
Y by MarkBcc168, jisunxingfu, A-Thought-Of-God, centslordm, megarnie, jhu08, HWenslawski, ImSh95, Adventure10, Mango247, Spiritpalm, deplasmanyollari, Akacool, Sedro, Korean_fish_Kaohsiung, Funcshun840, ItsBesi
Let 
denote the set of all positive rational numbers. Determine all functions 
satisfying 
for all 
denote the set of all positive rational numbers. Determine all functions 
satisfying 
for all 
This post has been edited 1 time. Last edited by Functional, Jul 17, 2019, 12:12 PM
.
for all 
.
denote the given statement, plug in 
to deduce 
.
finishes. 
![\[ f(y) = \left[ \frac{f(xf(y))}{f(x)} \right]^2. \]](http://latex.artofproblemsolving.com/8/d/f/8df379c9b7d6a6fdaca466edbfccf0fa0edb7bde.png)
We prove by induction on 
than all elements of 
are 
'
.
is vacuous, and the inductive step follows by above identity. It follows we must have 
equal to 
works if 
is replaced by 
.
gives us 
.
gives us 
or 
is a square of a rational for all positive rationals 
.
to get 
which again gives 
is a square of a rational for all positive rationals 
is a rational which is false for large 
is the only solution which works. 
.
,
for all positive rational numbers 
,
here means 
-
for every positive integer 
,
such that 
and we obtain 
![$\Rightarrow f^k(x)=\sqrt[k]{f(x)}$](http://latex.artofproblemsolving.com/0/6/0/060830f5be7bf7cc2a456995c900a3136546e5f8.png)
where 
.
such that 
let 
such that 
for all primes ![$f^k(a)=\sqrt[k]{f(a)}$](http://latex.artofproblemsolving.com/e/6/e/e6e0b5a0bd7795bce0b25221405e5c719e8c660c.png)
so ![$\sqrt[k]{f(a)} \in \mathbb{Q^+}$](http://latex.artofproblemsolving.com/4/4/c/44cc679356041331540a2361db34cc82a820dad5.png)
![$\sqrt[k]{\displaystyle\frac{m}{n}}=\displaystyle\frac{x}{y}$](http://latex.artofproblemsolving.com/2/0/6/206499cc50bba1a8cee02d0b02db68d2beb34a79.png)
for some 
with 
if 
Since 
this is clearly a contradiction and thus there doesn't exist any positive rational number 
with 
works. It clearly does. As usual, let 
.
If 
,
.
If we denote 
by 
then (*) implies 
.
,
,
,
,
for all 
Swapping 
and equating gives:
for some constant 
(as this holds 
)
for all 
.
.
:
and expanding out shows the result is true by induction.
for all 
Hence 
so we must have 
hence:
which indeed works
.![\[P(x,1)\implies f((xk)^2)=f(x)^2k.\]](http://latex.artofproblemsolving.com/5/6/8/56859dcc6a8e60e6b11b93e5de09553d82d2e9ae.png)
Plugging in 
into this, we get that 
,
.![\[P(x,1/k)\implies f(x^2)=f(x)^2.\]](http://latex.artofproblemsolving.com/d/c/0/dc0b089dd7d5ffd77805372327cca73324428db7.png)
Thus, the FE implies![\[f(y)=\left(\frac{f(xf(y))}{f(x)}\right)^2.\]](http://latex.artofproblemsolving.com/f/5/9/f596461ac3faa691dcece4fd6eb1d4f46a48aff7.png)
t
is trivial. Now, the equation we derived above shows that![\[f(y)=a^{2^{k+1}}\]](http://latex.artofproblemsolving.com/2/7/7/2775141a4b299ebd3d16f3d1d89701920d126bc0.png)
for some rational 
which is a rational number when taken 
-
at the given assertion, then 
.
for any 
,
for any ![$g(x), \sqrt{g(x)}, \cdots, \sqrt[2^k]{g(x)}, \cdots$](http://latex.artofproblemsolving.com/7/5/e/75e5b4cf85692c16b5bb7853061d5323cef79e44.png)
are all rationals, giving that 
is constantly 
and 
.
to get 
.
.
.
,
.
,
(
)
(
is rational). Now repeat the drill to get 
is rational 
.
Only solution is 
,
,
denote the initial assertion. From 
,
Now, from 
,
while from 
,
Hence, 
,
,
.
.
implies
From 
,
Therefore, if 
.
is in the image of 
.
.
.
.
.
.
.
.
.
is a 
power of a rational for all 
.![$f^{[n]}(x) = \underbrace{f(f(\dots f(x))\dots )}_{n \ times}$](http://latex.artofproblemsolving.com/c/d/0/cd0cd1051ca1e3f15e8ea73743c97be424edcfad.png)
we see that ![$$f^{[n+1]}(l) = f^{[n]}(f(l))= (f(f(l))^{\frac{1}{2^{n-1}}}=(f(l))^{\frac{1}{2^n}}$$](http://latex.artofproblemsolving.com/0/7/6/076ec9c7a9ac756fa1d86eeefcc088e30f9d578e.png)
And so our claim is proved.
.
denote the assertion and let 
denote the range of 
gives 
.
where 
gives ![\[f(xf(y)) = f(x) \sqrt{f(y)},\]](http://latex.artofproblemsolving.com/1/a/b/1abb090178c0ee8884bcbd327655250e4c0e15cf.png)
and letting 
.
implies 
.
,
is a perfect 
.
gives 
Since 
.
such that 
over the same domain and range. We have 
Taking the 
Since 
t
gives 
.
we get 
.
gives 
,
is in the range of 
so 
is also, meaning eventually 
.
which can be easily verified to work.
giving 
.
and the result follows.
This implies that 
power then 
power. Thus the only solution is 
to get that 
to get that 
to get that 
,
to get 
to get 
![$f(\sqrt{b})=\sqrt[4]{b}$](http://latex.artofproblemsolving.com/4/0/5/40501116d9bea9bfcbfba2ad128b038e1bb0507f.png)
,
for all 
,
and 
:![\[f(1) = f\left( \frac{1}{f(1)^2}\right)f(1)\]](http://latex.artofproblemsolving.com/9/c/8/9c8a7461fb6e9d18ff2b9bb720e3f491bd04c904.png)
![\[\implies f\left( \frac{1}{f(1)^2}\right) = 1.\]](http://latex.artofproblemsolving.com/b/e/3/be39a0e15efade08916efd76f5d3e2a2dc326c98.png)
gives![\[f(x^2) = f(x)^2.\]](http://latex.artofproblemsolving.com/b/8/f/b8f858d7c31ec613b8c74ab44f5afa29efc82218.png)
![\[f(xf(y))^2 = f(x)^2f(y)\]](http://latex.artofproblemsolving.com/7/f/a/7fa43484bfed75c0f2be53d643357ad0544b7ccc.png)
,![\[f(f(y))^2 = f(y).\]](http://latex.artofproblemsolving.com/4/a/c/4acc2c5db338cca7c5b17ee2a446a39deb2d81c0.png)
is in the range of 
,
is also in the range of 
,
,
.
,
.
for at least one prime 
is in the range of 
,
,
![\[f(\frac{1}{f(1)})=1\]](http://latex.artofproblemsolving.com/0/6/3/06393a2a089bb40b8556c42c8a77654763dfdba1.png)
Let 
.
![\[f(x^2)=f(x)^2\]](http://latex.artofproblemsolving.com/6/d/3/6d3e09d54ef8eb62b9a5d55a56568a7032acaedf.png)
Thus we get that 
,
.
gives us that 
gives us that 
gives us that 
,
.
,
,
,
.
;
.
,
and 
are 
-
.
and 
give us that 
.
such that 
.
.
with 
.
and so on.
and 
Then
However, note that 
thus 
Hence 
Thus 
so 
Then the equation reduces to 
and we can continue this infinitely. But 
unless it is 
Therefore 
is the only solution.
Since 
we can conclude that 
Taking 
into this equation implies 
![$\sqrt[2^k]{f(x)} \in \mathbb Q_{> 0}.$](http://latex.artofproblemsolving.com/d/6/a/d6a03e242189621acff50c67cb1a0a18d20f689b.png)
.
for some 
gives us ![\[\left( \frac{f(xf(y))}{f(x)} \right)^2 = f(y).\]](http://latex.artofproblemsolving.com/c/d/1/cd1051055085468468efc3b0cec0d5d7ed17fe36.png)
This implies that 
leads to 
.
such that 
.
,![\[
f(x)^2 = f(x^2) \quad (1)
\]](http://latex.artofproblemsolving.com/d/a/7/da744d6ab53144cbcd96c19cf701b9d124c215eb.png)
Using (1) in the original equation, we get:![\[
f(x f(y)) = f(x) f(\sqrt{y})
\]](http://latex.artofproblemsolving.com/3/d/0/3d0c04814da95ee990cc4f95b70b3d3e778480b0.png)
Setting 
and using symmetry, we derive:![\[
f(f(x)) f(\sqrt{y}) = f(f(y)) f(\sqrt{x})
\]](http://latex.artofproblemsolving.com/8/0/0/800a40c858435705c7c8b7bd4c6aaa9860776549.png)
From this, we conclude that:![\[
f(f(x^2)) = c f(x) \quad (2)
\]](http://latex.artofproblemsolving.com/a/9/9/a99bb4101583a5136f19bdd18a81ad0a3fcf34b2.png)
Now, applying 
to both sides of the original equation, we obtain:![\[
f(f((x f(y))^2)) = f(f(x)^2 f(y))
\]](http://latex.artofproblemsolving.com/2/5/0/250fb5c06fc4367ff423859fc67af3d827d9ce6d.png)
Using (2), this simplifies to:![\[
c f(x f(y)) = c f(f(x) \sqrt{f(y)})
\]](http://latex.artofproblemsolving.com/5/c/3/5c3ec69e794879de2fbb71a568fb2398a59af4e0.png)
Setting 
,
.![\[
f(f(x^2)) = c f(x) = f(x^2) = f(x)^2
\]](http://latex.artofproblemsolving.com/2/c/6/2c62cab83d9c0e108f64fc2ad7bb29c33ea4bcdf.png)
From this, it follows that 
,
.
we get 
which tells us that 
.
gives us 
which gives us 
Now notice that if I plug in 
,
but that means 
