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 5 of 8: The decisive

2

-adic elimination

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In plain words

Exact powers of $2$ in the difference equation rule out every larger multiple and force the sum $a+b$ to be exactly $2^{n-1}$ .

4\mid(c+1),\qquad v_2(c-1)=1,\qquad a+b=2^{n-1}

Detailed analysis

The right side of $(c+1)(b-a)=2^p-2^n=2^n(2^{p-n}-1)$ has exact $2$ -adic valuation $n$ . If $c$ were even, then $c+1$ would be odd, so $2^n\mid(b-a)$ , contradicting $0<b-a<2^{n-1}$ . Hence $c$ is odd. If $c\equiv1\pmod4$ , then $v_2(c+1)=1$ ; exact valuation $n$ would give $2^{n-1}\mid(b-a)$ , again contradicting $b-a<2^{n-1}$ . Therefore $c\equiv3\pmod4$ , so $4\mid(c+1)$ and $v_2(c-1)=1$ . Now $(c-1)(a+b)=2^n+2^p=2^n(1+2^{p-n})$ has exact valuation $n$ , hence $v_2(a+b)=n-1$ . Since $a+b<2^n$ , the positive integer $a+b$ must equal $2^{n-1}$ . Finally, if $a\ge5$ , then $a+b<2^{n+1}/(a-1)\le2^{n-1}$ , with strict inequality from $a+b<2c$ , contradicting $a+b=2^{n-1}$ . Thus $a\le4$ ; together with $a\ge3$ , only $a=3,4$ remain.