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I'd really appreciate help with this problem, because I'm stuck with it.

$$ \left\{ \begin{array}{c} ab+ac+ad+ae=-1 \\ ba+bc+bd+be=-1 \\ ca+cb+cd+ce=-1 \\ da+db+dc+de=-1 \\ ea+eb+ec+ed=-1 \end{array} \right. $$

I've tried substituting $a+b+c+d+e$ with t but it didn't get me far. I've also come up with equations $(a-b)(c+d+e)=0$ or $(b-c)(a+d+e)=0$.

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These equations $(a+b)(c+d+e)=0$.. seem promising. Can you add how did you get them and exactly what? And then, try what if $a+b=0$ and what if $c+d+e=0$.. –  Berci Sep 29 '12 at 15:47
    
there was a mistake, it's fixed now. I got those equations by subtracting one equation from another ( ab+ac+ad+ae-ba-bc-bd-be=-1-(-1) =>ac+ad+ae-bc-bd-be=(a-b)(c+d+e)=0 ) –  suomynona Sep 29 '12 at 15:54
    
I don't think there are any solutions. –  i. m. soloveichik Sep 29 '12 at 16:44

2 Answers 2

up vote 2 down vote accepted

Let $t=a+b+c+d+e$. Then your equations can be transformed to $a(t-a)=b(t-b)=\ldots=e(t-e)=-1$. From this $0=a(t-a)-b(t-b)=(a-b)(t-a-b)$ and similarly for all other pairs in place of $a,b$ as you found out yourself. We conclude that $a=b$ or $a+b=t$ (and equally for other pairs).

If $a=b=c=d=e$, this leads to $4a^2 = -1$, hence one solution is $$\tag1 a=b=c=d=e=\pm\frac12 i.$$

For the rest we may assume that not all variables are equal, wlog. assume $a\ne b$. Then $a+b=t$. Since we cannot have both $a=c$ and $b=c$ we conclude $a+c=t$ or $b+c=t$, hence $c=a$ or $c=b$. All in all, we see that $a,b,c,d,e$ take only two values. Up to symmetry, we have one of the following cases: $a\ne b=c=d=e$ or $a=b\ne c=d=e$.

In the first case, $t=a+4b$, i.e. $4ab=-1$ and $ab+3b^2=-1$, hence $3b^2=-\frac34$, i.e. $b=\pm\frac{1}2i$ and $a=-\frac1{4b}=b$, contrary to the assumption $b\ne a$.

From now on assume $a=b\ne c=d=e$. Then $t=2a+3c$, i.e. $a^2+3ac=-1$ and $2ac+2c^2=-1$, hence $c=-\frac{1+a^2}{3a}$ (note that clearly $a\ne0$), and by inserting $\frac{-4 a^4 - 2 a^2 + 2}{9 a^2}=-1 $. From this we find $-4a^4+7a^2+2=0$, hence $a^2=2$ or $a^2=-\frac14$. Hence we find solutions by letting (using $a^2=2$) $$\tag 2 \mathrm{Two\ numbers\ are\ }\pm\sqrt 2\mathrm{,\ the\ other\ three\ are\ }\mp\frac{\sqrt2}{2}.$$ On the other hand $a^2=-\frac14$ leads to $c=a$, hence no additional solution.

In total we have thus two solutions from $(1)$ and twenty solutions from $(2)$.

Note that only the solutions of type $(2)$ are real.

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D'oh, you should never derive your results live in $\LaTeX$ edit mode. I tried here and made a lort of mistakes that are hopefully gone now. –  Hagen von Eitzen Sep 29 '12 at 17:01
    
Thank you for the answer! –  suomynona Sep 29 '12 at 17:11

You are on the right track. So, all 5 of these is given: $(a-b)(c+d+e)=0$ (and rotating $a\to b\to c\to d\to e\to a$). If the 2nd member is never $0$, it means all the first members have to be $0$, i.e. $a=b=c=d=e$, and then they equal to $\displaystyle\pm\frac i2$.

If one of the sums become 0, say $c+d+e=0$, then from the first equation we get $ab=-1$, and can reduce the system to 3 variables..

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Thank you for the answer, but I can't see how it can be reduced to three variables... Could you please explain a bit? –  suomynona Sep 29 '12 at 16:30
    
Substitute $e=-c-d$ and $b=-1/a$ in the equations 3-5. But basically the other answer went into details. –  Berci Sep 29 '12 at 16:47

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