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2019 Greece Team Selection Test
4
(y^2+1)f(x)-yf(xy)=yf(x/y) for x,y>0
(y^2+1)f(x)-yf(xy)=yf(x/y) for x,y>0
Source: Greece TST 2019 p4
June 7, 2020
functional equation
functional
algebra
Problem Statement
Find all functions
f
:
(
0
,
∞
)
↦
R
f:(0,\infty)\mapsto\mathbb{R}
f
:
(
0
,
∞
)
↦
R
such that
(
y
2
+
1
)
f
(
x
)
−
y
f
(
x
y
)
=
y
f
(
x
y
)
,
\displaystyle{(y^2+1)f(x)-yf(xy)=yf\left(\frac{x}{y}\right),}
(
y
2
+
1
)
f
(
x
)
−
y
f
(
x
y
)
=
y
f
(
y
x
)
,
for every
x
,
y
>
0
x,y>0
x
,
y
>
0
.
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