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 6 of 8: The case

a=3

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In plain words

The forced sum $a+b=2^{n-1}$ turns the three equations into one difference of powers of $2$ .

a=3\quad\Longrightarrow\quad(a,b,c)=(3,5,7)

Detailed analysis

For $a=3$ , the identity $a+b=2^{n-1}$ gives $b=2^{n-1}-3$ . Using $3b-c=2^m$ and $3c-b=2^n$ , eliminate $c$ to obtain $2^n=2^m+8$ . If $m<3$ , division by $2^m$ would make an odd number equal to an even number; hence $m\ge3$ . Dividing by $8$ gives $2^{n-3}-2^{m-3}=1$ , so the smaller power is $1$ : $m=3$ , and then $n=4$ . Consequently $b=5$ and $c=(2^n+b)/3=7$ . The third value is $bc-a=35-3=32=2^5$ , so $(3,5,7)$ works.