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Solution to puzzle 68: Difference of powers

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


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

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

Set a = 2c, so that 32c − 2b = 1.
Then 2b = 32c − 1 = (3c − 1)(3c + 1).
By the Fundamental Theorem of Arithmetic, both (3c − 1) and (3c + 1) are powers of 2.
In fact, we must have 3c − 1 = 2 and 3c + 1 = 4, so that c = 1, and then a = 2.
Therefore the only solution for b greater than or equal to 3 is (a,b) = (2,3).

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

Source: Traditional

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