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5
f(n+2)-2022 \cdot f(n+1)+2021 \cdot f(n)=0
f(n+2)-2022 \cdot f(n+1)+2021 \cdot f(n)=0
Source: Moldova TST 2022
April 1, 2022
function
algebra
functional equation
Problem Statement
The function
f
:
N
→
N
f:\mathbb{N} \rightarrow \mathbb{N}
f
:
N
→
N
verifies:
1
)
f
(
n
+
2
)
−
2022
⋅
f
(
n
+
1
)
+
2021
⋅
f
(
n
)
=
0
,
∀
n
∈
N
;
1) f(n+2)-2022 \cdot f(n+1)+2021 \cdot f(n)=0, \forall n \in \mathbb{N};
1
)
f
(
n
+
2
)
−
2022
⋅
f
(
n
+
1
)
+
2021
⋅
f
(
n
)
=
0
,
∀
n
∈
N
;
2
)
f
(
2
0
22
)
=
f
(
2
2
20
)
;
2) f(20^{22})=f(22^{20});
2
)
f
(
2
0
22
)
=
f
(
2
2
20
)
;
3
)
f
(
2021
)
=
2022
3) f(2021)=2022
3
)
f
(
2021
)
=
2022
. Find all possible values of
f
(
2022
)
f(2022)
f
(
2022
)
.
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