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TST Round 1
2023 Taiwan TST Round 1
5
Divisible FE
Divisible FE
Source: 2023 Taiwan TST Round 1 Mock Exam P5
July 11, 2023
number theory
Taiwan
Problem Statement
Find all
f
:
N
→
N
f:\mathbb{N}\to\mathbb{N}
f
:
N
→
N
satisfying that for all
m
,
n
∈
N
m,n\in\mathbb{N}
m
,
n
∈
N
, the nonnegative integer
∣
f
(
m
+
n
)
−
f
(
m
)
∣
|f(m+n)-f(m)|
∣
f
(
m
+
n
)
−
f
(
m
)
∣
is a divisor of
f
(
n
)
f(n)
f
(
n
)
.Proposed by usjl
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