,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
to the side 
and 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
Y by Davi-8191, integrated_JRC, Vietjung, HWenslawski, Adventure10, megarnie, mathmax12, jmiao, aidan0626, Amir Hossein, buddyram, ItsBesi, and 3 other users
Find all functions 
(so 
is a function from the positive real numbers) such that
![\[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]](//latex.artofproblemsolving.com/0/4/b/04b955fb570ad384fb7f0e85bf0dacccf4c982a1.png)
for all positive real numbers
satisfying 
Author: Hojoo Lee, South Korea
(so 
is a function from the positive real numbers) such that![\[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]](http://latex.artofproblemsolving.com/0/4/b/04b955fb570ad384fb7f0e85bf0dacccf4c982a1.png)
for all positive real numbers
satisfying 
Author: Hojoo Lee, South Korea
This post has been edited 4 times. Last edited by MellowMelon, Sep 3, 2015, 12:05 AM
Reason: fix typo
Reason: fix typo
and observe that 
.
and consider 
.
independent of p.
and solve the quadratic equation resulting.
,
gives
so 
and you can get 
taking one of the values 
or 
at each value of 
)
,
![\[ \frac {f(s)}{s} = \frac { f(p) + f(\frac {s^{2}}{p}) }{p + \frac {s^{2}}{p}}
\]](http://latex.artofproblemsolving.com/2/7/1/2715a6c16d0a54da93a27dbb1f53414796b11b07.png)
with 
?
then we have![\[ f(x^2) = f(x)^2 \Rightarrow f(1) = 1
\]](http://latex.artofproblemsolving.com/b/5/5/b55d46c1a2319b0dae99d3ef5f87cb13dc528234.png)
we have :![\[ (y + z)(f(x) + f(w)) = (x + w)(f(z) + f(y))
\]](http://latex.artofproblemsolving.com/3/1/d/31d4488130aeb735170051e5d45d4dd26a39983a.png)
then we have :![\[ (y + z)f(\sqrt {yz}) = \sqrt {yz}(f(x) + f(y))
\]](http://latex.artofproblemsolving.com/d/6/c/d6c21c1b9bfba6c5b48f9b82a0ffb30016ba9d7f.png)
then :![\[ \frac {f(x)}{x} = \frac {f(x)^2)}{x^2 + 1} \Leftrightarrow (f(x) - x)(x.f(x) - 1) = 0 \Leftrightarrow f(x) = x,f(x) = \frac {1}{x}
\]](http://latex.artofproblemsolving.com/0/5/c/05c81ede5b26bbbe9dfe5f449ca3ec951b698490.png)
is different from 1 such that 
then
then![\[ (a + b)(f(ab) + 1) = (ab + 1)(a + \frac {1}{b})
\]](http://latex.artofproblemsolving.com/8/8/b/88b41c24e4c6ebf999aac3cb299e75dcce147700.png)
then
then :![\[ (a + b) = a(ab + 1) = b(ab + 1)
\]](http://latex.artofproblemsolving.com/1/1/0/1108b42155b7143b0362a3d3bf6f8eb9c48ea2e8.png)
(contradiction )
s
that mean:
we have : 
wich include
=
or 
to 
for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example).
for 
for 
being somewhere 
and somewhere 
.
.
with 
.
,
.
we have that
,
and 
.
we have 
for all 
we have 
.
,
depends only on 
,
.
.
and replacing 
by 
we get the quadratic equation in 
.
so that 
and 
,
.
(*). Now plug 
and we get 
(
is a strictly increasing function in 
we obtain that 
or 
.
(
,
and using the fact that 
we can assume that 
and 
are small enough or large enough to find that 
and 
(pluging 
)
one gets that 
)
![\[ \frac {( f(w) )^2 + ( f(x) )^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]](http://latex.artofproblemsolving.com/9/f/b/9fb83995e5477fb9e37277a09bb46c131a1ee044.png)
to find that 
.![\[ \frac {1 + f(x)^2}{2f(x)} = \frac {1 + x^2}{2x}
\]](http://latex.artofproblemsolving.com/9/d/b/9dbc46fc36620e7c2303913496023aecfe0da4f2.png)
.![\[ y + \frac {1}{y} = x + \frac {1}{x}\iff x - y = \frac {x - y}{xy}\implies x - y = 0\text{ or }xy = 1
\]](http://latex.artofproblemsolving.com/e/3/e/e3e92d98100eea7e6b7642aa6c2cb188f12bef3a.png)
![\[ f(x)\in{x,\frac {1}{x}}
\]](http://latex.artofproblemsolving.com/5/a/0/5a071f5bcd37d6c5f7363b4dfbbfd3649c2f49d2.png)
with![\[ f(a) = a,f(b) = \frac {1}{b}
\]](http://latex.artofproblemsolving.com/5/f/c/5fc852b080a66028fd3cd955a21e480639b565f8.png)
![\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a + b}
\]](http://latex.artofproblemsolving.com/b/4/b/b4b7dcc687acbf5231d9dddcf748c93b45c888f6.png)
,
can be one of two values:
,![\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a^2b + a} = \frac {ab + 1}{a + b}
\]](http://latex.artofproblemsolving.com/a/d/c/adcc211cd1f022cd5771d8d96557915dc3bef3e5.png)
,
,![\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a + \frac {1}{b}} = \frac {ab + 1}{a + b}
\]](http://latex.artofproblemsolving.com/f/9/a/f9a0dcb5f31c49654df05deef2e71bdec05000b5.png)
,![\[ f(x)\equiv x\text{ or }f(x)\equiv\frac {1}{x}
\]](http://latex.artofproblemsolving.com/4/7/a/47af025156362d5f11d6f113d0db5e58b99e5c97.png)
.
to show that 
.
.
)
,
then 
which is clearly only possible if 
or 
,
and 
are solutions to the FE and we want to prove their uniqueness.
![\[ \frac {\left( f(w) \right) + \left( f(x) \right)}{f(y) + f(z) } = \frac {w + x}{y + z}\]](http://latex.artofproblemsolving.com/5/e/2/5e2c11b7b81085f9440c4c854fc005793c147280.png)
for 
and solving the quadratic eqn. we have 
or 
none of them is 
satisfying 
. By considering 
we have a contradiction.
or 
,
yields ![\[f(x)^2 = f(x^2)\text{.}\]](http://latex.artofproblemsolving.com/5/9/5/595f969e40026e94f7a86f033abfccec15c7ea07.png)
This implies that 
.
for some positive real number 
Therefore, for each 
or 
.
,
and 
.
If 
,
which is clearly impossible. Similarly, if 
,
which is also impossible.
,
,
,
,
by the quadratic formula
,
denote the given assertion. From 
,
,
This rearranges to
Thus, for all 
,
and not equal to 
and 
.
,
Taking cases on whether 
or 
we see that:
,
,
,
,
,
,
,
for all 
and 
.
,![\[\frac{2f(c)^2}{2f(c^2)}=1 \implies f(c)^2=f(c^2),\]](http://latex.artofproblemsolving.com/d/1/c/d1cc9062b34804e209230466bbe30414cd2787fa.png)
since 
is over the positive reals. Additionally, we can also plug in 
to get that![\[f(1)^2=f(1) \implies f(1)=1,\]](http://latex.artofproblemsolving.com/9/5/7/9579a6d999b8018b69e40b1c7cd5c6fa44dac2b9.png)
again since 
,![\[\frac{f(\sqrt{c})^2+f(\sqrt{c})^2}{f(1)+f(c^2)}=\frac{c+c}{1+c^2} \implies \frac{2f(c)^2}{f(1)+f(c)^2}=\frac{2c}{1+c^2},\]](http://latex.artofproblemsolving.com/3/d/6/3d6d90a403d821ea5750dbe44209545a19ecec3c.png)
since we found by (1) that 
is fixed, letting 
,![\[\frac{k}{1+k^2}=\frac{c}{1+c^2},\]](http://latex.artofproblemsolving.com/d/5/9/d593b85878c930f81e39a694b6161f658df0456e.png)
![\[\iff k(1+c^2)=c(1+k^2),\]](http://latex.artofproblemsolving.com/f/1/8/f189539b72a40f01dfae9d8d8f4b9cd2c98e10aa.png)
![\[\iff ck^2-(1+c^2)k+c=0,\]](http://latex.artofproblemsolving.com/1/6/2/1627832bf7c8a075932214b974460b040f01557a.png)
![\[\iff (ck-1)(k-c)=0.\]](http://latex.artofproblemsolving.com/9/a/c/9acd10c5a385f41ee4effa4669263d1876812bc9.png)
This means that 
or 
,
,
is either 
,
such that 
.
then gives us that![\[\frac{f(ab)+f(ab)}{f(a)^2+f(b)^2}=\frac{ab+ab}{a^2+b^2} \implies \frac{f(ab)}{a^2+\frac{1}{b^2}}=\frac{ab}{a^2+b^2}, \]](http://latex.artofproblemsolving.com/a/b/1/ab1d9106a11b5e152842840f95b6d609290fb79a.png)
![\[\iff f(ab)=\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right).\]](http://latex.artofproblemsolving.com/f/9/a/f9a260fba39ba8101081531cdd4058bd501fede6.png)
By (2), we know that 
or 
.![\[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=ab,\]](http://latex.artofproblemsolving.com/2/7/9/279b1635713526c0b481fcc83742f4bca3e0577e.png)
![\[\iff \frac{a^2b^2+1}{b^2(a^2+b^2)}=1,\]](http://latex.artofproblemsolving.com/6/4/a/64a5cf3634c95c94ab98899b8176dbad1d83f862.png)
![\[\iff a^2b^2+1=b^2(a^2+b^2) \implies a^2b^2+1=a^2b^2+b^4,\]](http://latex.artofproblemsolving.com/c/b/a/cbae236b0c5c77e20dc15fa37f40011c22856154.png)
![\[\iff b^4=1,\]](http://latex.artofproblemsolving.com/0/b/b/0bbe49da89e8e8c2a59e4f6a2f75b796e2dbbae5.png)
![\[\implies b=1,\]](http://latex.artofproblemsolving.com/7/c/9/7c984e7e67f6833d59b909f9e6efa08e0ebf4bb3.png)
since 
,
.![\[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=\frac{1}{ab},\]](http://latex.artofproblemsolving.com/1/a/6/1a68d54c3dc7e6fd4f32e880a43fd17459a2ce60.png)
![\[\iff ab(ab)(a^2b^2+1)=b^2(a^2+b^2),\]](http://latex.artofproblemsolving.com/1/c/4/1c46c14ed0d6f485369bbb1c2c02e6f3c6a9a388.png)
![\[\iff a^4b^4+a^2b^2=a^2b^2+b^4,\]](http://latex.artofproblemsolving.com/9/1/0/910b5930c5964da5e7bb66d3d69d056ed77dab0f.png)
![\[\iff a^4b^4=b^4,\]](http://latex.artofproblemsolving.com/9/2/7/927a65b8a7803fc934cc6c7df8a336e2f37a7ae9.png)
![\[\implies a^4=1,\]](http://latex.artofproblemsolving.com/a/6/8/a68580ae2debae3d52db75e80592ea1d80fc8739.png)
![\[\implies a=1,\]](http://latex.artofproblemsolving.com/a/1/3/a132aef6ae7bb2216f859dd0d1c16097200092dc.png)
since 
or 
.
be the given assertion. Then 
gives 
gives ![\[
\frac{f(w^2)+f(x^2)}{f(y^2)+f(z^2)}=\frac{w^2+x^2}{y^2+z^2}
\]](http://latex.artofproblemsolving.com/a/0/2/a02463d2459d054962fc3adbf49d33e4eb894fe6.png)
whenever 
,![\[
\frac{f(a)+f(b)}{f(c)+f(d)}=\frac{a+b}{c+d}\tag{*}
\]](http://latex.artofproblemsolving.com/7/3/7/73727fb84b1578267a259dcd76c279b1819e0742.png)
whenever 
.
gives![\[
\frac{2f(x)}{f(x^2)+1}=\frac{2x}{x^2+1}
\]](http://latex.artofproblemsolving.com/7/c/5/7c5f2eb14351ccc23cc01467249e51276c5bdd67.png)
so![\[
(x^2+1)f(x)=xf(x)^2+x.
\]](http://latex.artofproblemsolving.com/2/f/5/2f5a68d596c336105d6c021958211a370c107916.png)
Solving this quadratic yields 
.
such that 
and 
.
into 
gives![\[
\frac{m+\frac{1}{n}}{f(mn)+1}=\frac{m+n}{mn+1}.
\]](http://latex.artofproblemsolving.com/e/2/9/e29765b5c48f4961854943ce81eadda1baa43c90.png)
Note that 
.
,![\[
\frac{m+\frac{1}{n}}{mn+1}=\frac{m+n}{mn+1}
\]](http://latex.artofproblemsolving.com/5/1/6/51685d0ed598d57e71ba01b0723f2b0f69b1a3ab.png)
so 
,
,![\[
mn\cdot\frac{m+\frac{1}{n}}{1+mn}=\frac{m+n}{mn+1}
\]](http://latex.artofproblemsolving.com/2/2/f/22f9617d7536e4e83ff08bdb525b1fa6dd5388ca.png)
so 
,
gives us that
Also, 
for some real number 
gives us that
after solving. Now, examining all of the cases, we find that a pointwise trap cannot occur, so we are done.
,
,
,
,
,
,
with 
,
,
respectively, giving us ![\[\frac{f(w)+f(x)}{f(y)+f(z)}=\frac{w+x}{y+z}.\]](http://latex.artofproblemsolving.com/6/2/0/620ced066d89129787764f3b86fc008fe98cfff1.png)
Note that this substitution works with the condition 
,
,![\[\frac{f(1)+f(x^2)}{2f(x)}=\frac{1+f(x^2)}{2f(x)}=\frac{1+x^2}{2x}.\]](http://latex.artofproblemsolving.com/c/3/8/c38b9c6394a6a7beab5eae0e87c3199935a1dd79.png)
Clearing the denominators and factoring this, we get ![\[2(f(x)-x)(xf(x)-1)=0,\]](http://latex.artofproblemsolving.com/1/e/b/1eb08730769767fdd36e3af6aee38d085387e571.png)
which gives 
,
or 
.
,
as ![\[\frac{a+\frac1b}{1+f(ab)}=\frac{a+b}{1+ab},\]](http://latex.artofproblemsolving.com/4/7/8/4785b164044ad2dd40d629fc8f305f40b530f9d0.png)
and by expanding and simplifying, it is easy to check that this equation does not hold for 
,
,
,
,
and we get 
.
,
,
.
OR 
for each individual 
.
there exist no values with 
,
for all 
,
,
.
satisfying 
,
,
,
we have 
,
.
,
and 
then we get 
.
,
then after a bit of simplification we get 
.
will give a contradiction (remember 
)
for all 
,
.
,
.
,
into our last result gives 
.
if 
into this gives 
or 
or 
.
such that 
,
.
for 
a contradiction. If 
then 
which is a contradiction. Therefore 
Also, the assertion then becomes 
Now, note that 
Therefore, 
so 
If there does not exist a 
then 
for all 
which is a valid solution. Otherwise, for 
in our domain with 
and 
If 
then 
and vice versa. If 
we have that 
a contradiction. Therefore 
It is easy to verify that they work.
which can easily be checked to work. Now we prove these are the only ones, setting 
we obtain 
for some constant 
and vice versa we obtain![\[\dfrac{f(w) + f(x) + 2c}{f(y) + f(z)} = \dfrac{w + x}{y + z} \]](http://latex.artofproblemsolving.com/3/a/8/3a8b0f32ea13ae55d75dad7c1e327fbe72f4164b.png)
for all 
.
gives 
then 
and 
for all 
yields![\[ \dfrac{f(w^2) + f(1)}{2 f(w)} = \dfrac{f(w)^2 + 1}{2 f(w)} = \dfrac{w^2 + 1}{2w} \implies f(w) \in \left \{w, \dfrac{1}{w} \right \} \forall w. \]](http://latex.artofproblemsolving.com/1/1/d/11d285ab8a24109059c2f55a8f4ed0d5959f2398.png)
From here we solve the pointwise trap, for the sake of contradiction assume 
exist where 
.
yields![\[\dfrac{f(a^2) + f(b^2)}{2 f(ab)} = \dfrac{a^2 + b^2}{2ab} \implies a(a^2b^2 + 1) = b(a^2 + b^2) f(ab) \]](http://latex.artofproblemsolving.com/0/4/9/049465ebf3e29e776881caaf005012e15c9ee6b6.png)
from 
and 
.![\[a(a^2b^2 + 1) = ab^2(a^2 + b^2) \implies a^3b^2 + a = a^3b^2 + ab^4 \implies b^4 = 1 \implies b = 1\]](http://latex.artofproblemsolving.com/c/9/c/c9c1f4dd50f0d79f951b3e1e7d248a46d0e46a92.png)
which fails. On the other hand, if 
we obtain![\[a^2b(a^2b^2 + 1) = b(a^2 + b^2) \implies a^4b^2 + a^2 = a^2 + b^2 \implies a^4 = 1 \implies a = 1 \]](http://latex.artofproblemsolving.com/d/9/6/d9648fa527e406dc84e7a9bf1aac506713b5c934.png)
which also fails. Contradiction, so 
gives us 
,
.
gives us 
.
and as we're going from the reals to the reals we must have 
and using quadratic formula on 
.
such that 
.
.
.
o
.
or 
we find that both of these give contradictions (we must have general 
