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Miklós Schweitzer
1995 Miklós Schweitzer
4
4
Part of
1995 Miklós Schweitzer
Problems
(1)
sum of powers
Source: miklos schweitzer 1995 q4
10/4/2021
For odd numbers
a
1
,
.
.
.
,
a
k
a_1 , ..., a_k
a
1
,
...
,
a
k
and even numbers
b
1
,
.
.
.
,
b
k
b_1 , ..., b_k
b
1
,
...
,
b
k
, we know that
∑
j
=
1
k
a
j
n
=
∑
j
=
1
k
b
j
n
\sum_ {j = 1}^k a_j^n = \sum_{j = 1}^k b_j^n
∑
j
=
1
k
a
j
n
=
∑
j
=
1
k
b
j
n
is satisfied for n = 1,2, ..., N. Prove that
k
≥
2
N
k\geq 2^N
k
≥
2
N
and that for
k
=
2
N
k = 2^N
k
=
2
N
there exists a solution
(
a
1
,
.
.
.
,
b
1
,
.
.
.
)
(a_1,...,b_1,...)
(
a
1
,
...
,
b
1
,
...
)
with the above properties.
algebra