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ITAMO
1999 ITAMO
6
6
Part of
1999 ITAMO
Problems
(1)
3^k -1 = x^n, 3^k -1 = x^3, diophantine
Source: 1999 ITAMO p6
1/25/2020
(a) Find all pairs
(
x
,
k
)
(x,k)
(
x
,
k
)
of positive integers such that
3
k
−
1
=
x
3
3^k -1 = x^3
3
k
−
1
=
x
3
. (b) Prove that if
n
>
1
n > 1
n
>
1
is an integer,
n
≠
3
n \ne 3
n
=
3
, then there are no pairs
(
x
,
k
)
(x,k)
(
x
,
k
)
of positive integers such that
3
k
−
1
=
x
n
3^k -1 = x^n
3
k
−
1
=
x
n
.
Diophantine equation
number theory
diophantine