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Dutch Mathematical Olympiad
1989 Dutch Mathematical Olympiad
5
(1+\sqrt2)^{2k+1}=n(k)+a(k)
(1+\sqrt2)^{2k+1}=n(k)+a(k)
Source: Netherlands - Dutch MO 1989 p5
December 25, 2022
algebra
Problem Statement
∀
k
∈
N
∃
n
(
k
)
∈
N
,
a
(
k
)
:
0
<
a
(
k
)
<
1
[
(
1
+
2
)
2
k
+
1
=
n
(
k
)
+
a
(
k
)
]
\forall k\in N \,\,\, \exists n(k) \in N, a(k):0<a(k)<1 [(1+\sqrt2)^{2k+1}=n(k)+a(k)]
∀
k
∈
N
∃
n
(
k
)
∈
N
,
a
(
k
)
:
0
<
a
(
k
)
<
1
[(
1
+
2
)
2
k
+
1
=
n
(
k
)
+
a
(
k
)]
Prove:
(
n
(
k
)
+
a
(
k
)
)
a
(
k
)
=
1
(n(k) + a(k))a(k) = 1
(
n
(
k
)
+
a
(
k
))
a
(
k
)
=
1
, for all
k
∈
N
k \in N
k
∈
N
, and calculate
lim
k
→
∞
a
(
k
)
\lim_{k \to \infty }a(k)
lim
k
→
∞
a
(
k
)
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