Solution for January 6, 2007

Find all natural numbers $a,b,c\,$ greater than one such that $(a-1)(b-1)(c-1)\,$ divides $abc-1\,$ exactly.

Proof by landen

I couldn't find a nice hand proof so I used the computer to help.

First result was that $(a,b,c) = (2,2,2),(2,2,4),(2,4,8),\,$ or $(3,5,15)\,$ are the only solutions for $a,b,\,$ and $c\,$ all less than $101.\,$

Let $k=\frac{abc-1}{(a-1)(b-1)(c-1)} = {{1-{{1}\over{a\,b\,c}}}\over{-{{1}\over{a\,b\,c}}+{{1}\over{b\,c}} +{{1}\over{a\,c}}-{{1}\over{c}}+{{1}\over{a\,b}}-{{1}\over{b}}-{{1 }\over{a}}+1}}$

If $k=2,\,$ then $(a,b,c)\,$ are all odd and $a\,$ is at least $3.\,$ $k\,$ is decreasing as $(b,c)\,$ increase. So for $a=3,\ (b,c)>(99,99)$ we have $k<1.54\,$ which is a contradiction. So for $k=2\,$ the only solution is $(3,5,15)\,$

If $a=2\,$ or $a=3\,$ we have $k=2\,$ for additional solutions by numerical calculation.

If $(a,b,c)>(4,4,4)\,$ then $k=2\,$ for additional solutions.

Since all $k=2\,$ solutions are known there are no others.

$\square$

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