Problem 2

Determine all triples

(a,b,c)

of positive integers such that each of the numbers

ab-c

bc-a

, and

ca-b

is a power of

2

(a power of

2

is an integer of the form

2^n

, where

n

is a nonnegative integer).

Step 8 of 8: Collecting and checking the solutions

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In plain words

The cases $a=2$ and $3\le a<b\le c$ exhaust the ordered possibilities, and symmetry restores all permutations.

(a,b,c)\in\{(2,2,2),(2,2,3),(2,6,11),(3,5,7)\}\ \text{up to permutation}

Detailed analysis

The preliminary step excluded $a=1$ and $a=b\ge3$ . Thus the ordered solutions are exactly $(2,2,2)$ , $(2,2,3)$ , $(2,6,11)$ , and $(3,5,7)$ . Direct substitution gives respectively the three values $(2,2,2)$ , $(1,4,4)$ , $(1,16,64)$ , and $(8,16,32)$ , all powers of $2$ . By the symmetry in the first step, these and only these triples, together with their permutations, are the complete answer.