Find the number of pairs $(a_1,a_2,…a_6)$ satisfying
$$a^2_1 +a^2_2 +···+a^2_6 =6$$
and
$$(a_1 +a_2 +···+a_6)(a^3_1 +a^3_2 +···+a^3_6)=36.$$
Work:
I squared the first equation and subtracted the second from it to get $2(a_1^2a_2^2 + a_1^2a_3^2 + \dots +a_5^2a_6^2) = a_1(a_2^3+a_3^3+…a_6^3) + a_3(a_1^3+a_3^3+ \dots + a_6^3) +\dots + a_6(a_1^3+a_2^3+ \dots + a_5^3).$
Problem: I am completely stuck from here.
Best Answer
We claim that there are no solutions. Suppose otherwise, that there is a solution. By the Cauchy-Schwarz Inequality, we have
$$ 36=(a_1 +a_2 +···+a_6)(a^3_1 +a^3_2 +···+a^3_6)\geq(a_1^2+a_2^2+\dots+a_6^2)^2=36 $$
Hence, equality for the Cauchy-Schwarz Inequality must be achieved. This means that $a_i^3/a_i=a_i^2$ must be constant, or that $|a_i|$ must be constant. Suppose that $|a_i|=x$. Then, $a_1, a_2, \dots, a_6$ must be either $x$ or $-x$. By the Pigeonhole Principle, there must be $3$ of $a_1, a_2, \dots, a_6$ that is equal. However, the condition requires $a_1, a_2, \dots, a_6$ to be distinct, and we have a contradiction. Hence, there are no solutions.