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2022 Czech-Austrian-Polish-Slovak Match
2
Fractional R+ FE
Fractional R+ FE
Source: Czech-Polish-Slovak Match 2022 P2
July 26, 2022
algebra
functional equation
Problem Statement
Find all functions
f
:
R
+
→
R
+
f: \mathbb{R^{+}} \rightarrow \mathbb {R^{+}}
f
:
R
+
→
R
+
such that
f
(
f
(
x
)
+
y
+
1
f
(
y
)
)
=
1
f
(
y
)
+
x
+
1
f(f(x)+\frac{y+1}{f(y)})=\frac{1}{f(y)}+x+1
f
(
f
(
x
)
+
f
(
y
)
y
+
1
)
=
f
(
y
)
1
+
x
+
1
for all
x
,
y
>
0
x, y>0
x
,
y
>
0
. Proposed by Dominik Burek, Poland
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