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Russian TST 2016
P2
Functional equation happens with a certain condition
Functional equation happens with a certain condition
Source: Russian TST 2016, Day 12 P2
April 19, 2023
algebra
functional equation
Problem Statement
Prove that a function
f
:
R
+
→
R
f:\mathbb{R}_+\to\mathbb{R}
f
:
R
+
→
R
satisfies
f
(
x
+
y
)
−
f
(
x
)
−
f
(
y
)
=
f
(
1
x
+
1
y
)
f(x+y)-f(x)-f(y)=f\left(\frac{1}{x}+\frac{1}{y}\right)
f
(
x
+
y
)
−
f
(
x
)
−
f
(
y
)
=
f
(
x
1
+
y
1
)
if and only if it satisfies
f
(
x
y
)
=
f
(
x
)
+
f
(
y
)
f(xy)=f(x)+f(y)
f
(
x
y
)
=
f
(
x
)
+
f
(
y
)
.
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