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 2 of 8: Preliminary exclusions

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In plain words

The smallest variable cannot be $1$ , and equality $a=b\ge3$ is incompatible with two neighboring powers of $2$ .

a>1,\qquad a=b\ge3\ \text{is impossible}

Detailed analysis

If $a=1$ , then $ab-c=b-c\le0$ , contradicting $ab-c=2^m\ge1$ ; hence $a>1$ . Since $b\le c$ and $a\ge2$ , we also have $2^n=ac-b\ge(a-1)c\ge2$ , so $n\ge1$ . Suppose $a=b\ge3$ . Then $ac-b=a(c-1)=2^n$ , so both $a$ and $c-1$ are powers of $2$ ; in particular $a$ is even and $c$ is odd. Thus $ab-c=a^2-c$ is odd, hence $a^2-c=1$ , so $c+1=a^2$ is a power of $2$ . But $c-1$ and $c+1$ are powers of $2$ differing by $2$ : writing them as $2^r$ and $2^s$ gives $2^r+2=2^s$ , which forces $r=1$ and $c=3$ . Then $a\le c$ gives $a=3$ , contradicting that $a$ is even. Therefore $a<b$ whenever $a\ge3$ .