Prove that there exist two powers of 3 whose difference is divisible by 1997.

Solution

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, 32,... 31997. It contains 1998 members. Therefore, by the Pigeonhole principle, some two of them, say 3n and 3m, n > m, have equal remainders when divided by 1997. Then their difference (3n - 3m) is divisible by 1997.

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