AoPS Academy
+1 w
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be the Euler's totient function.
, does there exist 
such that for any 
, 
and 
, 
is a non-negative power of 
?
, are there integers 
and 
satisfying:![\[
\varphi(k_i) \in \left ( \frac{x}{a} ,x \right ], i = 1,2; \quad \varphi(k_1) \neq \varphi(k_2).
\]](http://latex.artofproblemsolving.com/6/c/b/6cbe59f15b51439d1202267f521b573a52f1fe67.png)
And these two different 
correspond to the same 
and 
as described in (1), yet 
.
as the number of elements in set 
. For integer , let 
and 
. Compare 
with 
.
such that for all 
, we have 
.
such that, for all integers 
that satisfy 
, the following equality holds:![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
+1 w
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
, let 
be the 
-excenter. Points 
and 
are placed on line 
such that 
is between and 
, and is between and . Moreover, and are the contact points of with the -excircle of triangles 
and 
, respectively. Let 
be the -excenter of triangle 
, and let 
be the reflection of the orthocenter of triangle with respect to its circumcenter. Prove that , , and are collinear.
, let 
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer such that 
.
let 
be the foot of the altitude from 
to the side 
and 
, 
, 
be the incenter, -excenter, and 
-excenter, respectively. Denote by 
and 
the other intersection points of the circle 
with the lines 
and 
, respectively. Prove that 
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
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.
satisfying 
for all pairs of positive reals 
and 
.
denotes the set of all positive reals.
so there cannot exist a solution 
that is constant for sufficiently large 
.
then
.
then 
is injective. [Statement 2]
,
,
where 
is an integer...that does not make 
,
,
where 
,
,
,
where 
is any positive integer.
and any positive real 
,
.![$ [*]$](http://latex.artofproblemsolving.com/4/0/2/402cdd77190d3e8f32ee22b1bc288e7ee9044633.png)
for the positive integer 
.
are positive integers then
not in range of 
not in domain of 
for every positive 
such that 
,
- contradiction.
we have 
.
as msecco said. 
we have 
and if we fix 
and we assume 
for two positive numbers less than 
- impossible unless 
.
is injective and from this follows that 
is injective function, too.
,
.
,
and after substituting, we find 
.
is injective, which doesn't at all mean that 
satisfies the Cauchy equation and it's defined for the set of positive real numbers and gets only positive values. This doesn't mean that 
.
to 
on rationals and showing 
because 
and 
.
)
.
.
in (0) ![$ f(x + [f(y) - y]) = f(x) + f(y)\ \forall \ x > y\in \mathbb R^ +$](http://latex.artofproblemsolving.com/c/5/6/c565d15193f32ebfe852276c461ab043af811f05.png)
.
where 
.![$ f(f(y) + [f(z) - z]) = f(y + [f(z) - z]) + f(y)$](http://latex.artofproblemsolving.com/3/a/e/3ae94d81f27e3ce11b985be7c081228f5a711bda.png)
.![$ [f(f(y)) + f(z)] = [f(y) + f(z)] + f(y)\implies f(f(y)) = 2f(y)\ \forall \ y\in \mathbb R^ +$](http://latex.artofproblemsolving.com/b/4/5/b4576ab67341d72d75b61ff62cf6a3dd0b88d184.png)
.
in (2) and using (3) we have 
,
.
,
for some 
then putting 
in (0) and then putting 
in (0) we get 
.
but (6) gives 
.
,![\[ f(x+f(y))+z=f(x+y)+f(y)+z\]](http://latex.artofproblemsolving.com/d/2/3/d235165e929cda022e48788495537f2be30635a8.png)
![\[ f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)\]](http://latex.artofproblemsolving.com/0/1/1/01127e1b69a76ad9d3173b179a61b1bad7529150.png)
![\[ f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)\]](http://latex.artofproblemsolving.com/c/2/4/c245f0fe50133997f57a38293d76c52ba970ba37.png)
![\[ f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)\]](http://latex.artofproblemsolving.com/0/c/1/0c1279a16bc7328c3b6ca113cb6f396cabc88ff0.png)
![\[ f(x+y+z)+f(y)=f(x+2y+z)\]](http://latex.artofproblemsolving.com/a/8/5/a85c92d74cfce8b6a5bf9b37e5c820a0dcb3df1c.png)
sehingga ![\[ f(a)+f(b)=f(a+b)\]](http://latex.artofproblemsolving.com/4/6/6/4666abfa099267c77708cc0b0124e09776c89792.png)
and by Cauchy in positive reals, then 
for all 
.
,
for all positive real numbers 
gives 
,
gives 
.
,
,
,
.
.
of 
giving the same 
.
,
with arbitrarily large difference such that 
.
and 
,
,
.
.
for all 
.
denote the assertion to the functional equation.
and 
such that 
.
gives 
which is impossible for the given co-domain. Therefore 
.
such that 
.
and 
yield us that 
.
for periodicity. Choose a positive integer 
.![\[\alpha \le f(\alpha) < \alpha + mT \le f(\alpha + mT) = f(\alpha)\]](http://latex.artofproblemsolving.com/2/2/f/22ff5761b2128d5338a3974197c02d86e8e9e45c.png)
which is absurd. Hence the claim holds true. 
for all 
which is almost a Cauchy like condition.
and 
for 
for 
and 
to get
for 
and finally combine this with 
to get
for 
which is the partial Cauchy condition advertised above.
in 
.
.
to get![\[f(3x + f(x)) = f(4x) + f(x) = f(5x) \iff f(x) = 2x\]](http://latex.artofproblemsolving.com/1/c/a/1ca5c9b2014943287453005f5281f6002680f721.png)
due to injectivity. This completes the solution. 
,
we have 
.
for some 
,
by simply looking at the statement, which is not possible.
and get 
,
such that 
for all 
.
is![\[\forall x,y\in \mathbb R_+,\quad g(x+y+g(y))=g(x+y)+y\quad (*),\]](http://latex.artofproblemsolving.com/5/3/7/537f99a8ca79925e6ed27497b8285dd07fc58da9.png)
and we need to prove that 
for some reals 
.![\[g(x+y)+y=g(x+y+g(y))=g(x+y+g(x))=g(x+y)+x,\]](http://latex.artofproblemsolving.com/1/9/7/197d0c0abc1e41461f1fee25dd776f8a2d462e14.png)
i.e. 
.
.
as follows : For all real numbers 
,![\[g(A+g(y))=g(A)+y\quad (1)\]](http://latex.artofproblemsolving.com/a/1/3/a13272f2751c58ce991bad18b664290c2cdee694.png)
In particular, note that iterating gives![\[g(A+2g(y))=g(A+g(y)+g(y))=g(A+g(y))+g(y)\]](http://latex.artofproblemsolving.com/1/c/e/1ced97d66be65fcb9dc86b089547e503573a1de9.png)
We thus get by induction that 
for every positive integer 
.
and ![\[g(B+kg(y))=g(B)+ky=g(B+g(ky)),\]](http://latex.artofproblemsolving.com/d/8/d/d8d2c9df819728bd4d6007a756a90afd865f5c30.png)
where 
is some constant. By injectivity this implies that 
,
of 
takes all real values greater than 
takes values of 
,
such that 
,![\[y=\frac{t-g(A)}{k}<A.\]](http://latex.artofproblemsolving.com/a/8/2/a822681d66cc412f558aaa73d702f67bee685ccc.png)
We then have 
,
be such that 
.
.
By injectivity, we must have![\[ky+g(g(y_0))=(k+1)y_0\iff k(y-y_0)=T,\]](http://latex.artofproblemsolving.com/b/d/c/bdc43301bbc1b5ce8899d228f5c92dc4eec5dd66.png)
where we let 
.
large enough, thus it must be the case that 
,
.
.
.
,
and 
.
.
.
.
to be the plug-in of 
for all 
.
for some 
implies:
which is not possible, a contradiction. Thus the claim is proven.
such that 
.
Let 
:
Now assume the existence of 
such that 
.
and 
instead of 
This means 
in
However by injectivity, this means 
,
,
.
gives 
,
.
![\[f(x+f(y))=f(x+y)+f(y)\]](http://latex.artofproblemsolving.com/c/0/7/c07452c1395c2953830be7f6dc7bd80dd10fd03b.png)
We claim that 
gives![\[f(x+f(y))+f(x)=f(f(x)+f(y))=f(x)+f(y)+f(x+y)\]](http://latex.artofproblemsolving.com/f/5/4/f54cd8148b2e83fa41e5afcae5a10748fd817028.png)
This means that if 
,
.
gives 
.![\[f(f(x)+y)+f(y)=f(f(y)+x)+f(x)\Longrightarrow f(x)+2y=f(y)+2x\Longrightarrow f(x)-2x=f(y)-2y\]](http://latex.artofproblemsolving.com/d/0/5/d05e135e8836eb9d58ea8f6d0b910c72791ffdbe.png)
So, 
is a constant. So, 
.![\[2x+4y+3c=f(x+f(y))=f(x+y)+f(y)=2x+4y+2c\Longrightarrow c=0,\]](http://latex.artofproblemsolving.com/2/3/9/2395c83ff27389eb70af1d8864cdd99ea48e26dd.png)
and 
.
,
implies 
,
,
implies 
,
such that 
and 
gives that 
.
.
which is clearly impossible.
.
,
implies
(by (1)). (2)
.
in (3) and substracting (2) we get
.
in the original relation:
(by symmetry). (4)
in (4):
.
)
.
in this relation:
(using (5)) 
.
,
.
for any 
.
for some function 
If 
and 
,
for positive reals 
.
.
it is true, so assume 
.
.
,
,
and the output by 
such that 
.
.
to 
,
and this is valid as long as 
)
.
does not give 
but 
.
.
,
and 
for all 
by 
We wish to prove 
.
for all 
.
for all 
.
for some 
.
.
by 
.
,
for all 
,
.
yields![\[f(2x-f(x)) = 0,\]](http://latex.artofproblemsolving.com/e/4/1/e41e6a1ffe5046c704b4e2aa0557f493b2c2a279.png)
and this applies for all 
,![\[g(x+y+g(y)) = g(x+y)+y,\]](http://latex.artofproblemsolving.com/1/8/0/180292fe5606aea7c270d54b02a7697be3313324.png)
![\[g(a+g(b)) = g(a)+b,\]](http://latex.artofproblemsolving.com/3/f/e/3fecf2cbab637a37a8c47f59a8ad14d358cd3a0f.png)
.
.
and 
such that 
.
and 
yields![\[g(a)+k_1 = g(a+g(k_1)) = g(a+g(k_2)) = g(a)+k_2,\]](http://latex.artofproblemsolving.com/e/3/8/e38c8a3b20e4a3a285195ba39ed6957340deeae8.png)
gives![\[g(a+g(a+b)) = g(a)+(a+b) = g(a+g(b))+a = g(a+g(a)+g(b)),\]](http://latex.artofproblemsolving.com/a/7/d/a7dc5a12fed42d5ab193bf34c67968596f948079.png)
due to Claim 2. Since 
,
,
Thus it follows that 
for all 
.
which we get by putting 
in the equation. Now pick 
.
So we get 
which implies that 
.
holds for all 
.
for all 
be defined as ![\[g(z+g(y)) = g(z)+y, \qquad ( \diamondsuit )\]](http://latex.artofproblemsolving.com/f/6/7/f67d71cc2d330ee5b882722ebb5446799fd0695a.png)
where 
.![\[g(z + ng(y)) = g(z) + ny. \]](http://latex.artofproblemsolving.com/8/7/2/8722984fad092fb059aa427f631ba9f1bf2f8089.png)
So, we can vary the variables while fixing 
to get, due to injectivity, that![\[z+ng(y)=z+n'g(y') \implies ng(y) = n'g(y')\]](http://latex.artofproblemsolving.com/c/b/6/cb6b9a1068e67935810edcc5ffbc4a038c5fa9c0.png)
whenever 
.
for all positive reals 
for all 
for sufficiently large 
,![\[g(g(y)) = g((n+1)g(y)) - g(ng(y)) = y.\]](http://latex.artofproblemsolving.com/d/8/2/d8293011378fd700fd1b74da55f4799c4ee3c97c.png)
for all 
in 
(with 
gives us that![\[g(z+y) = g(z)+g(y)\]](http://latex.artofproblemsolving.com/2/6/5/26595c43c40d582af22dd572bcf5083c80e8b7b0.png)
for all 
.
,![\[g(x+y+g(y))=g(x+y)+y.\]](http://latex.artofproblemsolving.com/1/8/8/1887572dcd28f3aad2853b5b59cf04991ca5bf64.png)
Now setting 
we get ![\[g(x+g(y))+g(y)=g(x+g(y)+y)=g(x+y+g(y))=g(x+y)+y.\]](http://latex.artofproblemsolving.com/a/f/b/afb34696c77c7a7515d1b74e270ab5c64fd9dc7b.png)
gives ![\[g(y+g(y))=y+g(2y)-g(y).\]](http://latex.artofproblemsolving.com/2/3/5/2359e6ee3201be646ecc7a3450198f57a1caf4c3.png)
By quasi-additivity, 
so 
,
as well.
by Cauchy. Checking, we see 
,
and 
for some 
,
we get:![\[
f(2x)=f(2x)+f(x)
\]](http://latex.artofproblemsolving.com/2/d/b/2dbf72caf5eaef834cfcb7be66bd16d2fbb16b26.png)
However as 
this is not possible.
.
![\[f(x_0)=f(2x_0-f(x_0))+f(x_0)\]](http://latex.artofproblemsolving.com/2/f/4/2f4be9f9d7a6ce378f5cecbaf958078cc18512de.png)
This is a contradiction so 
.
denote:![\[g(x+g(y))=g(x)+y\]](http://latex.artofproblemsolving.com/4/f/6/4f6f460f8ce2e01cd317ed7348081d6eed08fa41.png)
This is equivalent to 
as we can let 
.
.
.
![\[g(y_0)=g(y_0-g(y_0))+y\]](http://latex.artofproblemsolving.com/a/0/4/a04e46d46bc3e193d8b2a642b83ab4975f1859bf.png)
Thus we must have 
.
and 
for 
.
![\[g(x_0+y_0+c)=x_0+y_0+k\]](http://latex.artofproblemsolving.com/e/9/5/e95bf7a539eea33b32303ee9ec80ec46cdd3c2fc.png)
However as 
this is a contradiction, thus 
and as 
we also must have 
.
to get 
which is a impossible, implying 
Define 
We get 
We also get 
Taking 
into our first equation, 
while we take 
to get 
which tells us that 
We have
Adding the first two and then subtracting the last, we get 
or 
or 
,
So 
when 
.
,![\[f(x)+f(x)+f(y)=f(x)+f(x+y)=f(2x+y)=f(2x)+f(y).\]](http://latex.artofproblemsolving.com/1/3/3/133ecf70e8fdd304a878e4b0d4c592cc06fb4b60.png)
Now 
so it is linear. Checking, only