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5
$m=1^{2k+1}+2^{2k+1}+\cdots+n^{2k+1}$
$m=1^{2k+1}+2^{2k+1}+\cdots+n^{2k+1}$
Source: Moldova TST 2023
April 8, 2023
number theory
Problem Statement
Find all pairs of positive integers
(
n
,
k
)
(n,k)
(
n
,
k
)
for which the number
m
=
1
2
k
+
1
+
2
2
k
+
1
+
⋯
+
n
2
k
+
1
m=1^{2k+1}+2^{2k+1}+\cdots+n^{2k+1}
m
=
1
2
k
+
1
+
2
2
k
+
1
+
⋯
+
n
2
k
+
1
is divisible by
n
+
2.
n+2.
n
+
2.
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