Dedicated to CONSTANTIN STELESCU on its 60th anniversary //Problem 111 Crux Mathematicorum v2n2 (Feb 1976)

      In the magazine

in Volume 2 Number 2 (February, 1976) page 25

     
     Two solutions of the exercise appear in Number 5 (May, 1976 page 95-96)

        Solution CiP

          The numbers $a,b,c$ can be considered as the roots of a third degree equation

$$x^{3}-px^{2}+qx-r=0.$$

 With the numbers $x_{1}=b-c,\;x_{2}=c-a,\;x_{3}=a-b$ we will form a third degree equation that they check. We will first calculate 

$$x_{1}+x_{2}+x_{3}=0$$

$$x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2} \cdot x_{3}=$$

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

$$=3\cdot q-p^{2}\overset{\underset{\mathrm{def}}{}}{=}Q$$

the last equality being in accordance with the Vieta's formulas,

$$x_{1} \cdot x_{2} \cdot x_{3}=(b-c)(c-a)(a-b)\overset{\underset{\mathrm{def}}{}}{=}R.$$

 Again with the Viète's formulas (for those who prefer the original French name François Viète instead of the Latinised form of his name, "Franciscus Vieta"), the numbers $b-c,\;c-a,\;a-b$ will be the roots of the equation

$$x^{3}-0\cdot x^{2}+Q\cdot x-R=0.$$

 But then $\frac{1}{x_{1}},\;\frac{1}{x_{2}},\;\frac{1}{x_{3}}$ will be the roots of the "written backwards" equation

$$R\cdot x^{3}-Q\cdot x^{2}-1=0$$

 and the Vieta's formulas give

$\sum_{cycl}\frac{1}{x_{1}} =\frac{Q}{R}$,            $\sum_{cycl} \frac{1}{x_{1}}\cdot \frac{1}{x_{2}}=0$.

     Now $\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}=\sum \frac {1}{x_{1}^{2}}=(\sum \frac {1}{x_{1}})^{2}-2 \cdot \sum \frac {1}{x_{1}\cdot x_{2}}=(\frac{Q}{R})^{2}-2\cdot 0=$

$$=(\sum_{cycl} \frac {1}{b-a})^{2}$$

that is, a square number.

$\blacksquare$