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Soros Olympiad in Mathematics
VI Soros Olympiad 1999 - 2000 (Russia)
9.1
k^{1999} - k^{1998} = 2k + 2 (VI Soros Olympiad 1990-00 R1 9.1)
k^{1999} - k^{1998} = 2k + 2 (VI Soros Olympiad 1990-00 R1 9.1)
Source:
May 27, 2024
number theory
Diophantine equation
Problem Statement
Prove that there is no natural number
k
k
k
such that
k
1999
ā
k
1998
=
2
k
+
2
k^{1999} - k^{1998} = 2k + 2
k
1999
ā
k
1998
=
2
k
+
2
.
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