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Moldova Team Selection Test
2006 Moldova Team Selection Test
1
Lucas divisibility
Lucas divisibility
Source: Moldova TST 2006, Test II problem 1
March 25, 2006
number theory solved
number theory
Problem Statement
Let
(
a
n
)
(a_n)
(
a
n
)
be the Lucas sequence:
a
0
=
2
,
a
1
=
1
,
a
n
+
1
=
a
n
+
a
n
−
1
a_0=2,a_1=1, a_{n+1}=a_n+a_{n-1}
a
0
=
2
,
a
1
=
1
,
a
n
+
1
=
a
n
+
a
n
−
1
for
n
≥
1
n\geq 1
n
≥
1
. Show that
a
59
a_{59}
a
59
divides
(
a
30
)
59
−
1
(a_{30})^{59}-1
(
a
30
)
59
−
1
.
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