Song's Complex System

$\frac{ab + bc + ca}{3} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$

$\frac{ab + bc + ca}{3} = (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2 - 2(\frac{ab+bc+ca}{abc})$

$\frac{ab + bc + ca}{3} = (\frac{ab + bc + ca}{abc})^2 - 2$

$\frac{ab + bc + ca}{3} = (ab + bc + ca)^2 - 2$

Let (ab + bc + ca) = x

$\frac{x}{3} = x^2 - 2$

$3x^2 - x - 6 = 0$

x= $\frac{1 \pm \sqrt{73}}{6}$

Therefore, $|x_1|+|x_2|= \frac{\sqrt{73}}{6}$

n=73 and m=3

Therefore, m+n=76

Suppose $ab=x,bc=y,ca=z$ .Then we have $abc(a+b+c)=ab \cdot bc+bc \cdot ca+ca \cdot ab=xy+yz+zx=1$ .From the second equation,we obtain :$$\frac{x+y+z}{3}=\frac{(ab)^2+(bc)^2+(ca)^2}{(abc)^2}$$ $$=x^2+y^2+z^2$$ $$=(x+y+z)^2-2(xy+yz+zx)$$ $$=(x+y+z)^2-2$$Thus,by substituting $x+y+z=a$ ,we have the quadratic equation:$$3a^2-a-6=0$$ $$|a 1|+|a 2|=\frac{\sqrt{73}+1}{6}+\frac{\sqrt{73}-1}{6}=\frac{\sqrt{73}}{3}$$ So, $n+m=73+3=76$

From the second equality, we can see that

$\frac{ab+bc+ca}{3} = \frac{1}{a^2} + \frac{1}{b^2} +\frac {1}{c^2}$

$= a^2b^2+b^2c^2+c^2a^2$ because $abc = (abc)^2 = 1$

$= (ab+bc+ca)^2 -2abc(a+b+c)$

$= (ab+bc+ca)^2 -2$ because $abc = a+b+c = 1$

let $ab+bc+ca = x$ so $\frac{x}{3} = x^2 -2$

so we have a quadratic equation $3x^2 - x - 6 = 0$

whose solution are $\frac{1 \pm \sqrt{1+4\times6\times3}}{6}$

$=\frac{1 \pm \sqrt{73}}{6}$

the absolute value of the solution are

$\frac{\sqrt{73}+1}{6}$ and $\frac{\sqrt{73}-1}{6}$

and the sum is $\frac{\sqrt{73}}{3}=\frac{\sqrt{n}}{m}$

so $m+n= 73+3=76$

For convenience, let $x=ab+ac+bc$ .

We need to solve for $x$ :

$\frac{x}{3}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

We will transform the RHS to get a function of only $x$ .

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2b^2c^2}\big(b^2c^2+a^2c^2+a^2b^2\big)=\big(*\big)$

Since $abc=1$ , this equals

$\big(*\big)=b^2c^2+a^2c^2+a^2b^2=\big(*\big)$

Now we use the identity $p^2+q^2+r^2=(p+q+r)^2-2pq-2pr-2qr$ with $p=bc, q=ac,r=ab$

$\big(*\big)=(bc+ac+ab)^2-2abc^2-2ab^2c-2a^2bc=\big(*\big)$

Because $abc=1$ , and also $a+b+c=1$ , this simplifies to

$\big(*\big)=(bc+ac+ab)^2-2(c+b+a)=x^2-2$ .

Now we have the following equation to solve:

$\frac{x}{3}=x^2-2$ , which has the roots $x_1=\frac{1+\sqrt{73}}{6}$ and $x_2=\frac{1-\sqrt{73}}{6}$

$|{x_1}|+|{x_2}|=\frac{\sqrt{73}}{3}$

Therefore, the solution is $76$ .

say ab+bc+ca = x and a^2b^2 + b^2c^2 + c^2a^2 = y

(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2+2(ab^2c+abc^2+a^2bc) = a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c) = a^2b^2+b^2c^2+c^2a^2+2 so x^2 = y+2

and

\frac{ab+bc+ca}{3} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2} = \frac{y}{1} = y So x = 3y y = \frac{x}{3} so x^2 = \frac{x}{3} + 2 3x^2-x-2 = 0 x = \frac{1 \pm \sqrt{73}}{6} only two possible solution so |x 1| + |x 2| = \frac{\sqrt{73}}{3} so m = 73 n = 3 m+n = 76

73+3