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 4 of 8: Bounds and parity in the remaining case

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In plain words

The middle power $2^n$ bounds $c$ , while the two largest equations force $a,b,c$ into a controlled parity pattern.

3\le a<b\le c\quad\Longrightarrow\quad c\le2^{n-1},\quad a+b<2^n,\quad 0<b-a<2^{n-1}

Detailed analysis

Assume $3\le a<b\le c$ . Since $2^n=ac-b\ge(a-1)c$ , we have $c\le2^n/(a-1)\le2^{n-1}$ . Also $a+b<2c$ , so $a+b<2c\le2^{n+1}/(a-1)\le2^n$ , and $0<b-a<c\le2^{n-1}$ . Because $n\ge3$ here, $2^n$ and $2^p$ are even. From $ac-b=2^n$ and $bc-a=2^p$ , if a is odd then b,c are both odd; if a is even then b is even. Thus a and b have the same parity, so $b-a$ is even. Adding and subtracting the two equations gives $(c-1)(a+b)=2^n+2^p$ and $(c+1)(b-a)=2^p-2^n$ . Since $b>a$ , the latter shows $p>n$ .