Question 1138427
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Given: {{{ab+ac+bc = abc}}}<br>
Move one of the terms on the left to the right side.  It doesn't matter which one, because the original equation is symmetric in a, b, and c.<br>
{{{abc-ab = ac+bc}}}
{{{ab(c-1) = c(a+b)}}}
{{{ab/(a+b) = c/(c-1)}}}<br>
That equation says that ab is an integer that is 1 more than (a+b).  So<br>
{{{ab = a+b+1}}}<br>
Now solve this equation for either variable in terms of the other.  Again it doesn't matter which, because again this equation is symmetric in a and b.<br>
{{{ab-a = b+1}}}
{{{a(b-1) = b+1}}}
{{{a = (b+1)/(b-1)}}}
{{{a = ((b-1)+2)/(b-1)}}}
{{{a = 1+2/(b-1)}}}<br>
1 is an integer, and a has to be an integer.  That means 2/(b-1) has to be an integer; and that means (b-1) has to be a factor of 2.<br>
So b-1 has to be either 1 or 2; that means b has to be either 2 or 3.<br>
And now we can find all the triples a, b, and c for which the given equation is true.<br>
(1) If b = 2 then<br>
{{{a = 1+2/1 = 1+2 = 3}}}<br>
and then<br>
{{{(ab)/(a+b) = c/(c-1) = 6/5}}} which means c is 6.<br>
The three numbers a, b, and c (in no particular order) are 2, 3, and 6.<br>
(2) If b = 3 then<br>
{{{a = 1+2/2 = 1+1 = 2}}}<br>
and then (as before)<br>
{{{(ab)/(a+b) = c/(c-1) = 6/5}}} which means c is 6.<br>
So there is a single set of three integers for which the given equation is true.<br>
Finally, since the problem specifies a < b < c, the single solution is<br>
{a,b,c) = (2,3,6)