AoPS Academy
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denote the assertion.
. Then from 
and 
we obtain 
, which holds for all 
. Thus let 
denote the set of all 
such that 
for all 
. Clearly 
, and is closed under multiplication and division. From 
we obtain 
, so 
, and therefore we have arbitrarily small (positive) elements of .
be arbitrary and compare with 
, which yields 
. Thus
For fixed 
and 
, by varying in 
the above fraction spans the interval 
. Since we can pick 
to be arbitrarily small it follows that all of 
is in , so then 
since 
, implying is constant. 
as the assertion to the functional equation.
give us that 
. If 
is necessarily injective, then we have 
which is impossible over . Therefore there exists distinct 
for which 
. and reveal there exists a for which 
for 
with 
.
and 
we get that![\[f(rx + f(1)) = f(x + f(1)) = f(rx + rf(1)) \implies f(x + f(1)) = f(x + rf(1))\]](http://latex.artofproblemsolving.com/b/8/9/b899b0ac342e1edfcf0bbe1355dfdef96e4c04bc.png)
which means is eventually periodic with period 
. Now compare and 
with large enough 
for periodicity to get![\[f(xy + yT) = f(xy).\]](http://latex.artofproblemsolving.com/2/f/a/2fa573094e6a91fe979d327174ae84799ecb834d.png)
Setting 
here would yield 
. This means is eventually constant for large enough domain values. Going back to the original equation set to be insanely huge. Then the RHS will be forced to be constant and the 
term will be constant as well. Simplifying the equation, we would get constant on as well. 
., then 
for all .
and 
. be the set of reals 
such 
for all . Note that is closed under multiplication.
it follows that ![\[ f(1 + f(x)) = f(1) \]](http://latex.artofproblemsolving.com/9/b/f/9bf23817bc729cf5d8c2b1a26cf0f0ef49cb7e8f.png)
so 
. As such, contains arbitarily large elements. is constant on 
. By 
it follows that ![\[ f(sx + f(y)) = f(x + f(y)) \]](http://latex.artofproblemsolving.com/3/b/e/3bea5701eb96f89451594dc3695c6c73dba1ce56.png)
so![\[ \frac{sx + f(y)}{x + f(y)} \in S \]](http://latex.artofproblemsolving.com/1/2/c/12c751efda2b7b4533f4ca7bf9d2b2be8f11a36c.png)
which has the image of interval 
as ranges. Taking arbitrarily large 
gives the result. 
for 
gives that is constant for all reals. Furthermore, all constant work.
and we get thatwhich means periodic with period is constant in 
. But since is periodic as well, this implies must be constant on as well. is periodic, but only on 
.
to prove that it works for 
too.
and be the assertion below 
.
Switch 
of the equation 
we have 
Let 
then 
then 
.
we have 
.
, 
And so 
., let some postive 
then 
.
then there exist 
such that 
and so 
. Thus the only solution is 
. Let's assume it is not constant., 
. and ,![\[ f(ax) = f(bx) \implies f(x) = f(rx), r=\frac{b}{a} > 1. \]](http://latex.artofproblemsolving.com/5/6/2/562850d6ae42843a2e6845ca99a386dd66ef38f9.png)
and 
.
.
for all 
. Now we can increase 
and we can increase indefinitely to cover the whole 
.
is a constant, contradiction. is injective.
. But putting 
, we get 
, contradiction again, and we are done. 
.
, where is a positive constant, which clearly works. gives 
. Hence, the function is clearly not injective.
such that 
. The above equations warrants the existence of at least few such pairs. Comparing 
and 
gives us that 
. It follows that if 
, we have for every 
. Let be the set of all positive real numbers for which . WLOG assume 
.
, then ![$[1,r]\in S$](http://latex.artofproblemsolving.com/0/9/b/09bde435a713191d6c4f0dcb542316023b419b0c.png)
. and to get that 
. So, for any 
we have 
. Let 
be a positive real. Then, we can set 
so that 
. The claim follows. 
for any positive integer 
, we have that 
and hence ![$[1,r^n]\in S$](http://latex.artofproblemsolving.com/f/5/0/f500639f2a0a387c09c82e6919c80aca77855e95.png)
for any . Since can be arbitrarily large, we have that 
. So, for any 
with 
, we have 
which means 
. It follows that the function is constant function. 
, which works. Denote the given assertion as . work, so we will suppose otherwise henceforth. is injective
and 
such that . Comparing and yields , so we have , where 
. Then, comparing and yields![\[f(x+f(y)) = f(rx+f(y)).\]](http://latex.artofproblemsolving.com/7/f/d/7fd1925816a584b69dc6ff99014a266fab43c9c4.png)
is periodic with period 
. Comparing and 
yields![\[f(xy) = f(xy+yp),\]](http://latex.artofproblemsolving.com/0/f/8/0f8f965bf5b9a29b9fd59afdfe95c139079997ad.png)
to make the period arbitrary. In other words, is constant, a contradiction. 
is injective, plugging in yields![\[f(f(1)+1) = f(1) \implies f(1) = 0,\]](http://latex.artofproblemsolving.com/b/1/8/b18f1c3ba29ac79972004a9bc000a0e12d5837b7.png)
exists, leaving only our solution set.
constant, which works.. Then 
give 
.
. Then gives 
. Thus 
.
for fixed give 
. Thus is periodic with period 
for 
.
for 
give 
so for any 
we have 
. Now if 
then 
for 
large enough so that 
. Thus is constant.
for some constant 
. These work. Now we show they are the only solutions. Suppose wasn't constant. is injective. for some 
. compared with gives , so 
, where 
. compared with 
gives![\[ f(x + f(y)) = f(cx + f(y))\]](http://latex.artofproblemsolving.com/e/2/2/e221d1d899be80d1d996ba65abb3c06ea1289a4c.png)
Thus, 
, so 
.
, we have 
.
so that for all 
, we have 
(possible by above).
, compared with 
gives that 
.
, varying gives that 
. Iterating this gives that 
and integers 
, which clearly implies is eventually constant. Suppose it is eventually constant at and 
.
, we have 
. Iterating gives that 
for any positive integer . Thus, for any 
, setting 
gives that 
, so is constant, a contradiction. gives 
, so 
, contradiction for 
.
and this tells us that is not injective. Let 
Taking 
we get that 
Let 
It is known 1 isn't the only element in . Take 
which gives 
Hence, 
and some fixed 
Taking 
and 
we get 
which implies 
This implies is constant.
. Comparing and gives that . So define the set such that if , then 
for all . Then 
.
gives that is not injective, so there is some with (this must exist if 
because if is in , so is 
). Then contains arbitrarily large elements (powers of ). Now comparing 
and 
where 
gives 
. So![\[\frac{saf(1)+f(1)}{af(1)+f(1)}=\frac{sa+1}{a+1}=s-\frac{s-1}{a+1}\in S.\]](http://latex.artofproblemsolving.com/a/3/7/a37574d153255301c70c77bb34e785d952b2222d.png)
This achieves everything between 
and exclusive, then reciprocating gives the other ones, so . So is just constant.
and 
Prove that
Where 
Where 
such that for all positive real numbers 
or 
,
for all real 
does not exist since 
which is given neither.
be the assertion of the equation.
and 
,
so that 
,
such that 
.
and 
to get 
.
,which gives that 
.
is constant!
but since 
is 
we have 
hence we get a solution 
![\[P(1, y)\]](http://latex.artofproblemsolving.com/9/7/6/9766cd2065738435858437b00850c537a4b0bb10.png)
![\[f(1+f(y))=f(1)\]](http://latex.artofproblemsolving.com/d/6/8/d685c18464262c12297d43ee65eea4a11e7865dc.png)
![\[P(x, 1+f(y))\]](http://latex.artofproblemsolving.com/d/9/1/d9150134b3736b924e65a1d4b21668010bc0080c.png)
![\[f(1)f(x+f(1))=f(1)f(x(1+f(y)))\]](http://latex.artofproblemsolving.com/7/d/0/7d0ea0aeb3a60fea534225e0ca9b756f5e5c79e7.png)
![\[P(x, 1)\]](http://latex.artofproblemsolving.com/1/1/b/11b17463c0a09adb906c249c4b957ce3e26f3d63.png)
![\[f(1)f(x+f(1))=f(1)f(x)\]](http://latex.artofproblemsolving.com/4/8/1/481f2d4d03cb2772ae9b0f9cfa7bbb54b4008ced.png)
Thus 
when 
![\[P(kx, y)\]](http://latex.artofproblemsolving.com/a/4/b/a4bcda851b16f81c96ba5691b6ed2225bf0382ac.png)
![\[f(y)f(kx+f(y))=f(y)f((xy))\]](http://latex.artofproblemsolving.com/4/f/8/4f86290a05f581fec6dcc1ce3653a98f9b31e0b1.png)
Thus 
thus let 
so 
.
for any ![\[P(x+c, y)\]](http://latex.artofproblemsolving.com/9/4/2/942efc9a5bb7c7e66cf27e4267392107e42d04f1.png)
![\[f(y)f(x+f(y))=f(y)f(xy+yc)\]](http://latex.artofproblemsolving.com/4/3/e/43e06fd3945c69627534924206e5aaa08012a9d8.png)
Thus 
for any