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Czech-Polish-Slovak Match
2018 Czech-Polish-Slovak Match
1
1
Part of
2018 Czech-Polish-Slovak Match
Problems
(1)
f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y) [Czech-Polish-Slovak Match 2018]
Source: Czech-Polish-Slovak Match 2018, Problem 1
7/2/2018
Determine all functions
f
:
R
ā
R
f : \mathbb R \to \mathbb R
f
:
R
ā
R
such that for all real numbers
x
x
x
and
y
y
y
,
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
Proposed by Walther Janous, Austria
function
functional equation
algebra