Prove that there exist two powers of 3 whose difference is divisible by 1997.
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Copyright © 19962018 Alexander Bogomolny
Prove that there exist two powers of 3 whose difference is divisible by 1997.
There are 1997 remainders of division by 1997. Consider a sequence of powers 1, 3, 3^{2},... 3^{1997}. It contains 1998 members. Therefore, by the Pigeonhole principle, some two of them, say 3^{n} and 3^{m}, n > m, have equal remainders when divided by 1997. Then their difference (3^{n}  3^{m}) is divisible by 1997.
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