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Problems
Contests
National and Regional Contests
Japan Contests
Japan TST
2022 Japan TST
2
Long R FE
Long R FE
Source: Japan TST 2022, Day 1, P2
September 17, 2022
algebra
functional equation
Problem Statement
Find all functions
f
:
R
→
R
f:\mathbb{R} \rightarrow \mathbb{R}
f
:
R
→
R
, such that
f
(
x
f
(
y
)
+
f
(
f
(
y
)
)
)
+
y
f
(
f
(
x
)
)
=
f
(
(
f
(
f
(
x
)
)
+
1
)
f
(
y
)
)
+
x
y
f(xf(y)+f(f(y)))+yf(f(x))=f((f(f(x))+1)f(y))+xy
f
(
x
f
(
y
)
+
f
(
f
(
y
)))
+
y
f
(
f
(
x
))
=
f
((
f
(
f
(
x
))
+
1
)
f
(
y
))
+
x
y
for all
x
,
y
∈
R
x, y \in \mathbb{R}
x
,
y
∈
R
.
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