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If you work it out you see that if

$a=b=c=1$, then $(1,1)=(1,1)=1$ and $1/1 + 1/1 + 1/1 = 3 =$ integer

and

$b=1, a=c=2$, then $(2,1)=(1,2)=1$ and $1/2 + 1/1 + 1/2 = 2=$ integer.

But please help me prove this

1 Answers 1

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So you have proven that the given cases does indeed work, and you need help to show that these are the only cases. So let's try to find another case. Let's see what the fraction sum turns out to be: $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc} $ Since $b$ have no factors in common with neither $a$ nor $c$, the last fraction cannot be shortened by any factor of $b$, so the denominator will always be a multiple of $b$, and thus the fraction cannot be an integer unless $b=1$.

If $b = 1$ (which we now know it must be for the conditions to hold), then $\frac{1}{a} + \frac{1}{c}$ must also be an integer, and thus none of $a$ and $c$ can be greater than $2$, and they have to be the same number. So either they are both $1$ or they are both $2$.

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    If $ \frac{bc + ac + ab}{abc} $ is an integer, then -- by multiplying with $ac$ -- so is $ \frac{bc + ac + ab}{b} = a + c + \frac{ac}{b}. $ Since $b$ have no factors in common with $ac$, if the last fraction is an integer, $b$ must be $1$.2012-12-10