Can anyone help me with a try on that problem? Thank you!
Let $F\in \mathbb{Z}\left [ X_{1},X_{2},X_{3},X_{4},X_{5} \right ], F=(X_{1}+X_{2}+X_{3})^{2}X_{4}X_{5}+X_{1}X_{2}X_{3}(X_{4}+X_{5})$. If $A = \{ a_1,a_2,a_3,a_4,a_5 \} \subseteq \mathbb{R}$ is a set with $|A| = 5$, find the maximum number of elements of the set: $$\{F(a_{\sigma(1)},a_{\sigma(2)},a_{\sigma(3)},a_{\sigma(4)},a_{\sigma(5)}) \} \mid \sigma \in S_5\}$$
$\text{My approach:}$ I think that's the ideea but i cannot redact the full problem: F is symetric with $X_1, X_2, X_3$ in relation with $X_4, X_5$ so i have maximum $10$ expressions and i have to show the founded number is $10$.
You have shown that there can be no more than ten elements. It remains only to find a set of ten.
Arbitrarily choose numbers $b_1,b_2,b_3,b_4,b_5$ so that no two of the $10$ pairs $\{ b_i,b_j\}$ have the same sum. Denote $\sum b_i$ by $B$ and let each $a_i=x-b_i$. Then $$ F(a_1,a_2,a_3,a_4,a_5)=11x^4-(8B+2b_4+2b_5)x^3+ \cdots$$
For the permutations $i,j,k,l,m$ of $1,2,3,4,5$, we therefore obtain ten different polynomials $F(a_i,a_j,a_k,a_l,a_m)$. Equating all pairs of these ten polynomials will only yield a finite number of roots and so we can choose (infinitely many ) $x$ so that we have the required ten different elements.