Solution to puzzle 68: Difference of powers

Find all ordered pairs (a,b) of positive integers such that |3^a − 2^b| = 1.

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By inspection, we have (a,b) = (1,1), (1,2), (2,3).
These are the only solutions with b 3.
We shall show that there are no other solutions with b 3.

Consider 3^a − 2^b ±1 (modulo 8), with b 3.
Since b 3, 2^b 0 (mod 8).
Further, 3² 1 (mod 8), and so 3²ⁿ 1 (mod 8), 3²ⁿ⁺¹ 3 (mod 8), for any non-negative integer, n.
Hence, if b 3, we must have 3^a − 2^b = +1, and a even.

Set a = 2c, so that 3^2c − 2^b = 1.
Then 2^b = 3^2c − 1 = (3^c − 1)(3^c + 1).
By the Fundamental Theorem of Arithmetic, both (3^c − 1) and (3^c + 1) are powers of 2.
In fact, we must have 3^c − 1 = 2 and 3^c + 1 = 4, so that c = 1, and then a = 2.
Therefore the only solution for b 3 is (a,b) = (2,3).

Hence the only ordered pairs of integers, (a,b), such that |3^a − 2^b| = 1, are (1,1), (1,2), (2,3).

Source: Traditional

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