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What can be the possible value of $a+b+c$ in the following case?

$$a^{2}-bc=3$$ $$b^{2}-ca=4$$ $$c^{2}-ab=5$$

$0, 1, -1$ or $1/2$?

After doing $II-I$, $III-I$ and $III-II$, I got, $$(a+b+c)(b-a)=1$$ $$(a+b+c)(c-a)=2$$ $$(a+b+c)(c-b)=1$$

I'm unable to solve further, please help.

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Try comparing the first two of your equations - $a+b+c \neq 0$ so $(c-a) = 2(b-a)$ and see if that helps you to make further progress. – Mark Bennet Jul 20 '12 at 6:08
up vote 6 down vote accepted

You have

$$\left\{ \begin{align*} &(a+b+c)(b-a)=1\\ &(a+b+c)(c-a)=2\\ &(a+b+c)(c-b)=1\;. \end{align*}\right.\tag{1}$$

These clearly imply that $a+b+c\ne 0$, so the first and third of these imply that $b-a=c-b$. In other words, $\langle a,b,c\rangle$ is an arithmetic progression. (The second equation of $(1)$ confirms this.) Set $d=b-a$; then $b=a+d$ and $c=a+2d$, so $a+b+c=3(a+d)$, and each of the equation in $(1)$ reduces to $3d(a+d)=1$.

Going back to the original equations, we see that

$$\begin{align*} 3&=a^2-bc=a^2-(a+d)(a+2d)\\ &=-3ad-2d^2=-3d(a+d)+d^2\\ &=d^2-1\;, \end{align*}$$

or $d^2=4$. Thus, $d=\pm 2$, $1=3d(a+d)=\pm6(a+d)$, and $a+b+c=3(a+d)=\pm\frac12$.

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maybe title of question has mistake,it could be possible values right?because if we get identities then ,it means that there is infinite solution right? – dato datuashvili Jul 20 '12 at 7:59
@dato: No, there are only two possible values of $a+b+c$. – Brian M. Scott Jul 20 '12 at 8:07
ok then i will delete my answer – dato datuashvili Jul 20 '12 at 8:10


Now $a-b=\dfrac1{a+b+c}$ etc.

Putting the values of $a-b, b-c, c-a;$



$$\implies a+b+c=\pm\frac12$$

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