We will look for a polynomial $p(x) = x^3 - a^2x$ and integers $a_i$ of the form $-2c(4c^2-a^2)$; then $p(x) + a_i = (x-2c)(x^2 + 2cx + 4c^2 - a^2)$ has roots $2c$ and $-c \pm \sqrt{a^2 - 3c^2}$, so we need $a^2 - 3c^2 = b^2$, and this for a fixed $a$ and $n$ different values for $c$ (and thus $b$). The general parametric solution for $a^2 - 3c^2 = b^2$ is $a=2(p^2+3q^2)$, $c=p^2-2pq-3q^2$, $b=p^2+6pq - 3q^2$, and it is a question to find $a$ that can be written as $a=2(p^2+3q^2)$ in $n$ different ways. But the set of numbers of the form $p^2 + 3q^2$ is closed under multiplication, since $(p_1^2 + 3q_1^2)(p_2^2 + 3q_2^2) = (p_1p_2 - 3q_1q_2)^2 + 3(p_1q_2 + p_2q_1)^2$. Thus it is possible to do just what we want.

Basically we're here asked to show existence of infinitely many triples $(a_i,b_i,c_i)$ such that $\sum a_i ,\sum a^2_i$ are constant but $\prod a_i$ is not. Taking $\sum a_i=0$ we need to show existence of infinitely many pair $(a,b)$ such that $a^2+ab+b^2$ if constant, while $ab(a+b)$ is not, taking $a,b\in\mathbb N$ if we get $(a+b)^2-ab=(x+y)^2-xy,ab(a+b)=xy(x+y)=t$ then $(a+b)^2-(x+y)^2=t(\frac{1}{a+b}-\frac{1}{x+y})\implies t<0$ ,absurd, so it's enough to existence of infinitely many pair $a,b$ such that $a^2+3b^2$ is constant.To do, take $a_k+i\sqrt3 b_k=(x+i\sqrt3 y)^{n-k}(p+i\sqrt3 q)^k$ and so $a_k-i\sqrt3 b_k=(x-i\sqrt3 y)^{n-k}(p-i\sqrt3 q)^k$ so, $a^2_k+3b^2_k=(x^2+3y^2)^{n-k}(p^2+3q^2)^k$ ,now so taking $x^2+3y^2=p^2+3q^2=15^2+3.3^2=12^2+3.6^2$ we get $a^2_k+3b^2_k=252^n$ for all $k\leq n$ so done.

Here is my solution (Hope it works) We consider the Pell type equation: $x^2-3y^2=1$ Now, since it has a solution $(2,1)$ it must have infinitely many solutions $(x,y)$={$(m_1,k_1),...,(m_n,k_n),...$} Now, let $y_i=\frac{2^{t_0}L}{m_i}$ where $t_0 > 10^{one trillion}$ (Lol.) where $L$ is the least common multiple of $m_1,...,m_n$ and $N=2^{t_0-1}L$. So, $(2N)^2-3x_i^2=y_i^2$ Letting $b=N^2$ we get that $4b=3x_i^2+y_i^2$. Now, consider $v_i=\frac{x_i^2-y_i^2}{4}$ and observe that $b+v_i=x_i^2$ $b-3v_i=y_i^2$ Letting $\alpha_i=\frac{(x_i+y_i)}{2}$ and $\beta_i=\frac{(x_i-y_i)}{2}$ and $\gamma_i=-x_i$ we conclude that $\alpha_i+\beta_i+\gamma_i=0$ $\alpha_i\beta_i+\beta_i\gamma_i+\gamma_i\alpha_i=-b$ and that $\alpha_i\beta_i\gamma_i=-y_i^3\frac{(k_i^2-1)}{2}k_i$ which is obviously a different number $\forall 1\le i\le n$ since, for a fixed $n$, since the given product decreases as $i$ increases by explicitly converting the general solution. Hence the constructed polynomial $p(x)=x^3-bx$ is an $n-good$ one. (Note we only need $t_0 > 10^{one trillion}$ for ensuring that the numbers defined $\forall 1\le i\le n$ are integers. So more or less it is a troll condition just for fun.)

The polynomial $p(x)=x^3-2x^2+x$ has the property that for any $k \in \mathbb{N}$, the roots of $$p(x)+\frac{-4(k^2-1)^2}{(k^2+3)^3} = 0$$are $\frac{4}{k^2+3}, \frac{(k-1)^2}{k^2+3},$ and $\frac{(k+1)^2}{k^2+3}$. The conclusion is now obvious.