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 MSTang, Davi-8191, anantmudgal09, bel.jad5, Ramanujan_math, Gerninza, itslumi, TheMathCruncher_007, samrocksnature, centslordm, HWenslawski, megarnie, Amoonguss, ImSh95, David-Vieta, Mogmog8, Adventure10, Mango247, seansean01347, aidan0626, and 4 other users
Find all function 
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](//latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where 
is greatest integer not greater than 
Proposed by Pierre Bornsztein, France
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](http://latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where 
is greatest integer not greater than 
Proposed by Pierre Bornsztein, France
This post has been edited 1 time. Last edited by djmathman, Apr 27, 2015, 2:58 PM
Reason: formatting
Reason: formatting
.
or 
.
If 
we get 
,
or 
,
.
we get 
or 
.
to find 
,
yields 
.
to the original we get 
.
we have 
,
or 
.
be the assertation of ![$f([x]y)=f(x)[f(y)]$](http://latex.artofproblemsolving.com/9/e/0/9e0ed442f0a777313bcd39334b9f8a9cfb9d8417.png)
![$P(0,0): f(0)=f(0)[f(0)]$](http://latex.artofproblemsolving.com/2/d/9/2d99307564e7e5237a88abf80ab03461c1707d21.png)
![$[f(0)]=1$](http://latex.artofproblemsolving.com/e/d/9/ed92e9c5461547a3607189c39bb5cf97b9c3b6eb.png)
.
for some constant 
.![$c=c[c]$](http://latex.artofproblemsolving.com/7/8/c/78c89a1ff9feada60a3f5f57a497e967b09aa2bb.png)
![$[c]=1$](http://latex.artofproblemsolving.com/c/7/d/c7decbf7aba841a2e7346dcf2ac08ca54e117ec8.png)
.
.![$P(1,1): f(1)=f(1)[f(1)]$](http://latex.artofproblemsolving.com/2/9/7/2976c35e4758ed1311ab383f2187c65081a36c2c.png)
.
is also a solution.![$[f(1)]=1$](http://latex.artofproblemsolving.com/8/e/4/8e4a94fe3e706e2d1f3200aa882ef44842bbde4b.png)
.![$P(x,1): f(x)=f([x])$](http://latex.artofproblemsolving.com/0/a/2/0a2b6f3354e596150e68dc05eb2369c3e8e28311.png)
.![$P(2010,\frac1{2010}): f(1)=f(2010)[f(\frac{1}{2010})]=f(2010)[f(0)]=0$](http://latex.artofproblemsolving.com/a/b/9/ab99ea8227edfca784b8754971f9b6cf36c45a6a.png)
.![$1=[f(1)]=[0]=0$](http://latex.artofproblemsolving.com/f/a/8/fa873ac948831e29700f67cd9eaa83a49575596d.png)
.
.
![$f(0) = f(x) [f(0)]$](http://latex.artofproblemsolving.com/f/f/a/ffa5de96bfbb448541eab5739b3c4cdeae1de588.png)
.![$[f0)] \ne 0$](http://latex.artofproblemsolving.com/d/7/7/d7793f0e973aa1f875ac4707c37834c3a6eb2a94.png)
then ![$f(x) = \frac{f(0)}{[f(0)]}$](http://latex.artofproblemsolving.com/1/f/f/1ffbf99a50b9d9764f0843c1c48440dad7c9ba03.png)
.
is constant. Let ![$c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$](http://latex.artofproblemsolving.com/f/9/5/f9525d83386947b586f5abf38e68a7c8b95fabd7.png)
,
or 
![$[f(0)] = 0$](http://latex.artofproblemsolving.com/c/6/f/c6fec13f046ad56994eabd8ad3ec02629594d269.png)
then ![$f(y)=f(1)[f(y)]$](http://latex.artofproblemsolving.com/1/5/3/153d39a03cd328633cfb538f754e6de8c51e0884.png)
.
then ![$[f(y)] = \frac{f(y)}{f(1)}$](http://latex.artofproblemsolving.com/5/8/a/58af81556dda20ce198f12dac4227831a691a308.png)
and substituting this in (1) we have ![$f([x]y)=\frac{f(x)f(y)}{f(1)}$](http://latex.artofproblemsolving.com/e/3/d/e3d990ca4a158c3b950324f98381b6432dfb8cf8.png)
.![$f([x]y)=f(x[y])$](http://latex.artofproblemsolving.com/8/9/4/8947c91ed36c468380c5270b38498d3f3e7fadae.png)
.
we get 
.
for all 
.
is a constant. Let 
,
.
is such that 
,
,
,
,
is not constant and equal to 
,
implies that 
.
is equal to 
,
and 
,
.
.
for all 
.
to get 
.
for all 
.
.
and we get 
,
,
for all 
.
and the equation becomes 
.
for all 
for all 
,
gives 
for all 
in the original equation. We get 
.
with 
,
and 
,
,
to get 
,
,
for all 
for all 
,
,
for some 
,
for all 
.
,
.
this implies 
.
;
,
,
for all 
constant, with 
,
![$f(1)=f(1)[f(1)], f(1)=0, [f(1)]=1$](http://latex.artofproblemsolving.com/8/f/d/8fd615b53795c362d83694850920254e8b1eb2b5.png)
and fix a. Then,![$f([a]y)=f(a)[f(y)]=f(a)[f(0)]$](http://latex.artofproblemsolving.com/c/0/e/c0e1d6c9bb194325a6897f06fad036e46665597c.png)
![$x\in [0, [a]]$](http://latex.artofproblemsolving.com/a/a/e/aae07dcbd8ec1837cef9c24f281901c11205ecf1.png)
(
or 
.
and the claim is satsfied. Let 
where 
Let 
.
Let 
.
Let 
.
.
.
.
where 
.
.
.
.
.
,
and 
.
,
and 
.
and the claim is satisfied.
we get :
or 
for all y
we get :
for all n then for 
we must have 
for all y so 
w
and for 
:
where 
and 
:
for 
for all y but 
so :
for all 
:
f
we get 
fo
f
f
f
-
is not zero and 
.
.
.
for all 
.
.
is not equal to 
.
.
for each 
for each 
is a number in 
.
s.t. 
is not zero.
.
.
.
to Q. then since 
we get 
.
will give a contradiction since 
is not zero.
![$f(1)=f(1)[f(1)]$](http://latex.artofproblemsolving.com/6/8/1/6819ee1fb1a592c46b6ed5328667cbac83eb1977.png)
so we have 
,![$f([x])=f(x)$](http://latex.artofproblemsolving.com/f/6/9/f69409a6d7649bb40a7d7915767b3988c3cd6c3b.png)
.
be a positive integer, and let 
and let 
be very small number. Then because of 
,![$f([k+\epsilon])=f([k-\epsilon])$](http://latex.artofproblemsolving.com/c/4/8/c48bf44a60e68e8aead7e206fd1b8db40ddd1e69.png)
.![$f([k+\epsilon])=f(k)$](http://latex.artofproblemsolving.com/b/8/1/b81602f0105b232de3653fa2ff8a8eb9de01aad9.png)
and ![$f([k-\epsilon])=f(k-1)$](http://latex.artofproblemsolving.com/1/4/9/149b2a93c0ddfbcd66e5b22c3734fedb817566fe.png)
.
,
,![$f([k+t])=f(k)$](http://latex.artofproblemsolving.com/7/c/1/7c1313c8c0826c74779df196d498ca8577976a00.png)
.
and hence the function 
.
where 
.
and 
where 
is a constant.
.
,
.
because 
as 
.
such that 
,
for all 
.
.
for all 
.
gives a contradiction.
.
,
for all 
.
,
to get that 
,
gives 
,
works, where 
gives 
,
gives 
gives 
.
gives 
.
is not equal to 
is equal to 1.( If 
is 1 .As a result, 
.
.
gives us that 
for that 
or any 
.
.
we have 
for any real 
,
such that 
.
,
.
,
then 
yields 
such that 
and 
(choose any 
such that 
has the same sign as 
)
we conclude that 
,
.
we then get 
.
![$f(x) = c \forall c \in [1, 2]$](http://latex.artofproblemsolving.com/e/5/f/e5fafa55cdddedfa2c9c4a6d98bd6539cde404e7.png)
.
to get 
which implies 
.
then plug in ![$f(0) = f(t) = c \in [1, 2]$](http://latex.artofproblemsolving.com/7/f/e/7fe422b2ab83fabcf9e86e43b1f5d5fa0b555d64.png)
.
to get 
.
gives 
then 
.
to get 
,
.
,
.
.
.
,
.
.
to find that 
can be any real number, 
.
.
.
or 
.
gives 
,
.
.
or 
We use proof by contradiction.
gives 
since 
.
.
implies 
,
for all 
,
,
.
,
gives 
for all ![\[f(\lfloor x\rfloor) = f(x)\lfloor f(1)\rfloor = f(x).\]](http://latex.artofproblemsolving.com/3/2/1/3213658ab36b04aa29c2d02ef80b0ca0a431e4fe.png)
Then, for any 
,![\[f(\lfloor x\rfloor y) = f(x)\lfloor f(y)\rfloor = f(x) \lfloor f(\lfloor y\rfloor)\rfloor = f(x)\cdot 0 = 0,\]](http://latex.artofproblemsolving.com/8/c/9/8c91e8a6233bef89c2ce356958cea93f239c8759.png)
so since for any 
and let 
,
,
is that 
gives us 
Now if 
then this gives us 
where 
now 
define a number 
Now
and subsituting 
with 
gives us 
and give is 
Then for all 
we have that 
If there is some 
we have that 
Otherwise, for all 
we have 
Then 
Thus 
so 
Clearly 
![$f([x]y) = f(x)[f(y)]$](http://latex.artofproblemsolving.com/7/a/6/7a6b59c8145b1b4c33f656d3ce95ede3079d31aa.png)
.![$f(0) = f(0)[f(0)]$](http://latex.artofproblemsolving.com/4/d/d/4dd3c28eceab935d557e9080ea826370fd14a6cb.png)
.![$[f(0)] = 1$](http://latex.artofproblemsolving.com/8/4/5/845243bcd07bfa9cc01e9a1a05f15005e196cc10.png)
![$[c]$](http://latex.artofproblemsolving.com/5/2/b/52bae999d8320de9922bb4da965c7d6eb93f7ed1.png)
should equal to 1. We can easily check that for any such ![$f(1) = f(1)[f(1)]$](http://latex.artofproblemsolving.com/f/f/2/ff2b35ace58c18634169b02cd4ba46e4a443dc3f.png)
.![$\forall x>2: P(x,\frac{1}{[x]})$](http://latex.artofproblemsolving.com/b/4/f/b4ffa6b0a19dba41684cb6a8509f1f89e95f4df7.png)
gives us ![$f(1) = f(x).[f(\frac{1}{[x]}]$](http://latex.artofproblemsolving.com/4/e/5/4e564d72371b93a3367d2585755995f77f1ba80b.png)
and as we know ![$[f(\frac{1}{[x]})] = [f([\frac{1}{[x]}])] = [f(0)] = 0$](http://latex.artofproblemsolving.com/6/4/0/640eae9ff73cbf56c4c36eb9391ce8364228cff5.png)
.
and 
.
,
.
.
.
.
,
.
.
,
,
then make 
and scale down by 
for all 
,
.
.
where 
,
.
yields 
.
.
implies 
.
gives![\[ f(1) = f(2) f \left(\dfrac{1}{2} \right) = f(2) f(0) = 0 \]](http://latex.artofproblemsolving.com/7/7/a/77a4bc30015126ca1a3470cc2ff5bb6c2f9b174c.png)
which fails.
to find that 
gives us 
.
we find 
where 
.
this gives us 
.
or 
we have 
.
covers every real number between 
where 
