Show that if $a,b,c$ are complex numbers that such that \[ (a+b)(a+c)=b \qquad (b+c)(b+a)=c \qquad (c+a)(c+b)=a\]then $a,b,c$ are real numbers.
Problem
Source: Romanian TST 2001
Tags: trigonometry, complex numbers, algebra proposed, algebra
spanferkel
17.01.2011 01:35
WakeUp wrote: Show that if $a,b,c$ are complex numbers that such that \[ (a+b)(a+c)=b\quad (1) \\ (b+c)(b+a)=c\quad (2) \\ (c+a)(c+b)=a\quad (3) \] then $a,b,c$ are real numbers. If $b=c$ then $a=b=c=0$ or $a=b=c=\frac 14$. Now suppose $a\ne b\ne c\ne a$. Substracting, we get $(a+b)(a-b)=b-c$. Multiplying the 3 cyclics, we find $(a+b)(b+c)(c+a)=1$ and so also $abc=1$. Further $q:=ab+bc+ac=b-a^2=c-b^2=a-c^2$, so $a^2-b^2=b(c-a)$. Combining with $a^2-b^2=b-c$, we find $c-a=1-\frac cb$, so $3=\sum_{cyc}\frac ba=\sum_{cyc}a^2b$. So from $(a+b)(b+c)(c+a)=1$ we get $\sum_{cyc}\frac ab=\sum_{cyc}ab^2=-4$. Thus $\frac ab,\frac bc,\frac ca$ are solutions of $x^3+4x^2+3x-1=0$. And these are three reals. It is easy to see then that $a,b,c $ must be reals too.
Letting $s:=a+b+c$, we find by adding all three $s^2+q=s$.
Multiply (1) by $a$ etc. and add up. So, summing this up, $\sum a^3=(1-s)q=s(1-s)^2$. Using $s^3=\sum a^3+3$ we thus get $\boxed{(s+1)(2s-3)=0}$.
If $s=-1$, then $q=-2$ and we get the well known system
\[a^2-b=2\ \ and \ cyclics,\] the roots of which are indeed real and moreover nice - see here. But only the roots $(a,b,c)=(2\cos\frac{2\pi} 7,2\cos\frac{4\pi} 7,2\cos\frac{6\pi} 7)$ and cyclics fit in the original system. Their quotients also solve $x^3+4x^2+3x-1=0$. So we don't need to investigate the other case $s=\frac 32$ any more.