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National and Regional Contests
Taiwan Contests
TST Round 2
2018 Taiwan TST Round 2
2
Functional equation 011
Functional equation 011
Source: 2018 Taiwan TST Round 2, Day 2, Problem 2
April 13, 2018
algebra
Problem Statement
Find all functions
f
:
Z
→
Z
f: \mathbb{Z} \to \mathbb{Z}
f
:
Z
→
Z
such that
f
(
x
+
f
(
y
)
)
f
(
y
+
f
(
x
)
)
=
(
2
x
+
f
(
y
−
x
)
)
(
2
y
+
f
(
x
−
y
)
)
f\left(x+f\left(y\right)\right)f\left(y+f\left(x\right)\right)=\left(2x+f\left(y-x\right)\right)\left(2y+f\left(x-y\right)\right)
f
(
x
+
f
(
y
)
)
f
(
y
+
f
(
x
)
)
=
(
2
x
+
f
(
y
−
x
)
)
(
2
y
+
f
(
x
−
y
)
)
holds for all integers
x
,
y
x,y
x
,
y
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