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The Question:

Let a,b,c be complex numbers satisfying $abc = 1$ and $a+b+c =$ $\frac1a + \frac1b + \frac 1c$ Show that at least one of $a,b,c$ must equal $1$.

What I have tried: Rearranging the $RHS$ and subbing in the first equation we get $a+b+c = bc + ac+ab$

Now from Equation 1 we have $ a = \frac{1}{bc}$ and subbing this into the manipulation above we get $\frac{1}{bc} + b+c = bc + \frac 1b + \frac 1c$

Now multiplying out by $bc$ we get $1+b+c = (bc)^2 +c + b$ implying that $(bc)^2 = 1$ And from equation one we get $a^2b^2c^2 = 1^2 = 1$ and $b^2c^2 = 1$ therefore $a^2 = 1$ and $a = 1$.

Is this correct/sufficient if not can you point me in the right direction and feel free to show other methods etc. Thanks.

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    Also note that, because you are working with complex numbers, you cannot assume, in this context, that square roots are positive.2012-12-29

2 Answers 2

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Let $s=a+b+c = bc + ac+ab$.

Then $a,b,c$ are the roots of

$(X-a)(X-b)(X-c)=X^3-(a+b+c)X^2+(ab+ac+bc)X-abc$ $=X^3-sX^2+sX-1$

Now all you have to do is observe that $X=1$ is a root of $X^3-sX^2+sX-1$.

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    I have always thought this an elegant way of looking at such problems - however it does not always work as well as this. One indications for using this approach is that all the given equations are symmetric.2012-12-29
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Alternatively, you can just write:

$(a-1)(b-1)(c-1)=abc - (ab+ac+bc) + (a+b+c) -1 = 1 - [ab+ac+bc-(a+b+c)] - 1 = 0$

So one of $a-1,b-1,c-1$ must be zero.