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3^k -1 = x^n, 3^k -1 = x^3, diophantine

Source: 1999 ITAMO p6

January 25, 2020
Diophantine equationnumber theorydiophantine

Problem Statement

(a) Find all pairs (x,k)(x,k) of positive integers such that 3k−1=x33^k -1 = x^3 . (b) Prove that if n>1n > 1 is an integer, n≠3n \ne 3, then there are no pairs (x,k)(x,k) of positive integers such that 3k−1=xn3^k -1 = x^n.
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