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 7 of 8: The case

a=4

is impossible

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In plain words

The ordering inequalities squeeze a power of $2$ strictly between $2$ and $3$ .

a=4\quad\Longrightarrow\quad 2^{n-3}\le3\ \text{and}\ 2^{n-3}>2

Detailed analysis

For $a=4$ , $a+b=2^{n-1}$ gives $b=2^{n-1}-4$ . The equation $4c-b=2^n$ then gives $c=3\cdot2^{n-3}-1$ . Since $b\le c$ , we get $2^{n-1}-4\le3\cdot2^{n-3}-1$ , hence $2^{n-3}\le3$ . Since $a<b$ , we also have $4<2^{n-1}-4$ , hence $2^{n-3}>2$ . No power of $2$ lies strictly between $2$ and $3$ , so $a=4$ yields no solution.