# Proof of three variable equation.

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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"Now multiplying out by $bc$ we get" Are you sure we get that? – Nameless Dec 29 '12 at 21:04
Ah, let me re-work thanks! – fosho Dec 29 '12 at 21:05
Also note that, because you are working with complex numbers, you cannot assume, in this context, that square roots are positive. – Mark Bennet Dec 29 '12 at 21:15

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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+1 This is the neat way that probably occurs to the more "mature" visitors who were taught all this stuff at school? – Old John Dec 29 '12 at 21:09
Yes, when you reach a certain level, all problems of symmetric polynomials suggest this sort of approach :) – Thomas Andrews Dec 29 '12 at 21:11
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. – Mark Bennet Dec 29 '12 at 21:12

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.

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