All permutations needed

Find the sum ( a + b + c ) (a+b+c) of all ordered triples of positive integers ( a , b , c ) (a,b,c) such that each of the following numbers a b c , b c a , c a b \large ab-c, \quad bc-a, \quad ca-b is a power of 2.


The answer is 231.

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1 solution

Bob Kadylo
Feb 25, 2018

16 such triplets exist: (2,2,2), 3 permutations of (2,2,3), 6 permutations of (2,6,11) and 6 permutations of (3,5,7)

6 + 3 times 7 + 6 times 19 + 6 times 15 = 231 \boxed {231} It took 30 minutes of Trial and Error.....

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