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Czech-Polish-Slovak Match
1998 Czech and Slovak Match
4
f (n)+ f (n+1)= f (n+2) +f (n+3) -168
f (n)+ f (n+1)= f (n+2) +f (n+3) -168
Source: Czech and Slovak Match 1998 P4
October 1, 2017
Natural Numbers
functional equation in N
algebra
Problem Statement
Find all functions
f
:
N
→
N
−
{
1
}
f : N\rightarrow N - \{1\}
f
:
N
→
N
−
{
1
}
satisfying
f
(
n
)
+
f
(
n
+
1
)
=
f
(
n
+
2
)
+
f
(
n
+
3
)
−
168
f (n)+ f (n+1)= f (n+2) +f (n+3) -168
f
(
n
)
+
f
(
n
+
1
)
=
f
(
n
+
2
)
+
f
(
n
+
3
)
−
168
for all
n
∈
N
n \in N
n
∈
N
.
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