A problem with Symmetric Polynomials // Um problema com polinómios simétricos

 It is problem 3 016, 2/1977 page 59 from the "Revista Matematica a Elevilor din Timisoara" (Mathematical Magazine of Students from Timisoara).

         "3 016.  If  $a,\;b,\;c$  are nonzero real numbers, such that 

 $a+b+c=0$  and  $a^3+b^3+c^3=a^5+b^5+c^5$ , 

then  $a^2+b^2+c^2=\frac{6}{5}.$

{author : } Titu ANDREESCU, student, Timisoara"

Solution CiP

                    We will use the following algebraic identities :

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

$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a) \tag{2}$

$(a+b+c)^5-a^5-b^5-c^5=$

$=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) \tag{3}$

          (1)  is obtained from the identity :

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

          For  (2)  let  $P(a,b,c)=(a+b+c)^3-a^3-b^3-c^3$. If  $b=-a$  then  $Pa,-a,c)=c^3-a^3+a^3-c^3=0$  so  $P$  has the factor  $a+b$. From symmetry it will also have the factors  $b+c,\;c+a$  so, being of degree 3, it results 

$P(a,b.c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\;deg(Q)=0$

hence  $Q(a,b,c)$  is a constant. As  $P(1,1,0)=2^3-1-1=6$, from  $6=(1+1)(1+0)(0+1)\cdot Q(1,1,0)$  it foloow   $Q=3.$

          For  (3)  let  $P(a,b,c)=(a+b+c)^5-a^5-b^5-c^5$.  $P(a,-a,c)=c^5-a^5+a^5-c^5=0$  so

$P(a,b,c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\deg(Q)=2 \tag{4}$

So  $ Q$  has the form  $Q(a,b,c)=m\cdot (a^2+b^2+c^2)+n\cdot (ab+bc+ca)$

If we put in  (4)  first  $a=b=1,\;c=0$,  and second  $a=b=c=1$  we get 

$P(1,1,0)=2^5-1-1=30=2(m\cdot 2+n),\;$

$\;\;P(1,1,1)=3^5-1-1-1=240=8(m\cdot 3+n\cdot 3)$

hence  $m=n=5$.

               Let's move on to solving the problem. In the condition  $a+b+c=0$ , we have :

 - from  (1)  

$\sum a^2=-2\sum ab \tag{5}$

  - from  (2)   $0-\sum a^3\;\underset{a+b=-c\;etc}{=}\;\;\;3(-c)(-a)(-b)$ , so

$\sum a^3=3abc \tag{6}$

  - from  (3)  $0-\sum a^5=5(-c)(-a)(-b)(\sum a^2+\sum ab)\overset{(5)}{=}-5abc(\sum a^2-\frac{1}{2}\sum a^2$c , so

$\sum a^5=\frac{5}{2}abc\sum a^2 \tag{7}$

But we also have  $\sum a^3=\sum a^5$ , so from  (6)  and  (7)  it results 

 $3abc=\frac{5}{2}abc \sum a^2$

hence the answer.

$\blacksquare$