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International Contests
Austrian-Polish
1984 Austrian-Polish Competition
9
9
Part of
1984 Austrian-Polish Competition
Problems
(1)
f (x + y) = f (x)f (y) - f(xy) + 1 for all x,y \in Q
Source: Austrian Polish 1984 APMC
4/30/2020
Find all functions
f
:
Q
→
R
f: Q \to R
f
:
Q
→
R
satisfying
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
−
f
(
x
y
)
+
1
f (x + y) = f (x)f (y) - f(xy) + 1
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
−
f
(
x
y
)
+
1
for all
x
,
y
∈
Q
x,y \in Q
x
,
y
∈
Q
functional
functional equation
algebra