MathDB
Problems
Contests
International Contests
Czech-Polish-Slovak Match
2009 Czech-Polish-Slovak Match
1
1
Part of
2009 Czech-Polish-Slovak Match
Problems
(1)
Functional equation: ( 1 + y f(x) )( 1 - y f(x+y) ) = 1
Source: Czech-Polish-Slovak Match, 2009
8/21/2011
Let
R
+
\mathbb{R}^+
R
+
denote the set of positive real numbers. Find all functions
f
:
R
+
→
R
+
f : \mathbb{R}^+\to\mathbb{R}^+
f
:
R
+
→
R
+
that satisfy
(
1
+
y
f
(
x
)
)
(
1
−
y
f
(
x
+
y
)
)
=
1
\Big(1+yf(x)\Big)\Big(1-yf(x+y)\Big)=1
(
1
+
y
f
(
x
)
)
(
1
−
y
f
(
x
+
y
)
)
=
1
for all
x
,
y
∈
R
+
x,y\in\mathbb{R}^+
x
,
y
∈
R
+
.
function
symmetry
algebra unsolved
algebra