BurLy also looking for "sanda" // "Gazeta Matematica" magazine, series B

 M-am pomenit aseară că și  "BurLy"  a auzit de  "sanda" , dar nu mai știa dacă nu cumva e... sandra...

Although I said I wouldn't get emotionally involved again when leafing through old magazines from my youth, here I cannot help but note the following.

1.  Complete Algebraic Identities

          On page 357, within the solution to  Problem 24444, the identity is mentioned without any proof or commentary :

$a^3+b^3+c^3+d^3-3(abc+abd+acd+bcd)=$

$=(a+b+c+d)(a^2+b^2+c^2+d^2-ab-ac-ad-bc-bc-cd) \tag{1}$

Let's explain a little. The following identity is well known :

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

We are trying to see if we have a  complete identity  of form :

$\sum a^3-\sum abc=(\sum a)\cdot (\sum a^2-\sum ab) \tag{3}$

Let us consider the polynomial involving the four letters  $a,\;b,\;c\;and\;d$

$P(a,b,c,d):=[\;a^3+b^3+c^3+d^3-3(abc+abd+acd+bcd)\;]-$

$-[\;(a+b+c+d)(a^2+b^2+c^2+d^2-ab-ac-ad-bc-bc-cd)\;]$

We have, for  $d=0$ ,

  $P(a,b,c,0)=[\sum a^3-3abc]-[(\sum a)(\sum a^2-\sum ab)]\underset{(2)}{=}0$

(Above, the sums $\sum$  extend to three letters.)

Therefore, the polynomial  $P$ is divisible by  $d$ , and by symmetry, it is also divisible by  $a$, $b$, and $c$, and thus by their product :

$P(a,b,c,d)=abcd\cdot Q(a,b,c,d) \tag{4}$

However, on the left-hand side of the equation  (4) , the degree of the polynomial is three, whereas on the right-hand side,  $abcd$  is of degree four. Therefore, in the equation  (4) $Q$  must be identically zero. This is how we find the identity  (1).

     So, we have seen that the identity  (3)  holds for four letters. If we add a fifth letter,  $e$, then the identity remains valid, in the same form. This is the meaning of the word "complete" mentioned earlier. The reasoning is the same :

$P_1(a,b,c,d,e):=[\sum a^3-3\sum abc]-[(\sum a)(\sum a^2-\sum ab)]_{e=0}\;\overset {(3)}{\underset{for\;a,b,c,d}{=}}\;0$

So  $P_1$  is divisible by  $e$ , therefore, by symmetry, it is also divisible by  $a,\; b,\; c,\; and \;d$. But in the writing

$P_1(a,b,c,d,e)=abcde\cdot Q_q(a,b,c,d,e)$

comparing the degrees of the two polynomials leads to the conclusion that  $Q1$  is identically zero.

           The fact that the formula  (3)  holds for any number of letters is expressed by some authors by stating that "the identity  (3)  is saturated, starting with three letters." However, the term "saturated" is not widely used. See

MIHĂILEANU N.[N.] Complemente de algebră elementară (Ed. Didactică și Pedagogică,

București, 1968)

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2.   Titu's Lemma  (revived)

The inequality that Titu Andreescu claims as his own appears again on page 323, in Lemma 1. It remains to be seen what the priority is for circulating this inequality. I would point out that the parenthetical citation of Lemma 1 is INCORRECT. You have RMT 1-2 /1979 here.