Is Vieta Useful?

Let the roots be $a,b,c$ . Then by Vieta's we have that

(i) $a + b + c = 0$ ,

(ii) $ab + ac + bc = -3$ and

(iii) $abc = -n$ .

Now from the first two of the conditions we have that

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + ac + bc) = 0 \Longrightarrow a^{2} + b^{2} + c^{2} = 6$ .

Now for three perfect squares to add to $6$ , two of them must be $1$ and one of them $4$ , so without loss of generality let $a = \pm 1, b = \pm 1$ and $c = \pm 2$ . To satisfy (i) we can either have $(a,b,c) = (1,1,-2)$ or $(a,b,c) = (-1,-1,2)$ , each of which satisfies (ii). The resulting products $abc$ are $-2$ and $2$ , respectively, yielding values for $n$ of $2$ and $-2$ . The desired product is then $2*(-2) = \boxed{-4}$ .

Let an integer root of the equation be $a$ , then the equation can be written as follows:

$\begin{aligned} x^3 -3x +n & = (x-a)(x^2+bx+c) \\ & = x^3 + (b-a)x^2 +(c-ab)x -ac \end{aligned}$

Equating the coefficients, we have:

$\begin{cases} b - a = 0 & \Rightarrow b = a \\ c-ab = c - a^2 = -3 & \Rightarrow c = a^2-3 \\ n = -ac & \Rightarrow n = 3a - a^3 \end{cases}$

For factor $x^2+bx+c$ to have integer roots its determinant: $b^2 - 4c = a^2 - 4a^2 + 12$ $= 12-3a^2$ $= m^2$ where $m$ is a non-negative integer.

The possible $a = \begin{cases} \pm 1 & \Rightarrow m = 3 & \Rightarrow n = \pm3 \mp 1 = \pm 2 \\ \pm 2 & \Rightarrow m = 0 & \Rightarrow n = \pm 6 \mp 8 = \mp 2 \end{cases}$ .

Therefore, the product of all $n$ is $-2(2) = \boxed{4}$ .

Just to show that Vieta's formulae are not necessary...

The cubic $f_n(x) = x^3 - 3x + n$ has derivative $3x^2 - 3$ , so has turning points at $\pm1$ . For $f_n(x)$ to have three integer roots, it must have three real roots, and hence $f_n(-1) \ge 0 \ge f_n(1)$ . This implies that $|n| \le 2$ .

If $|n| < 2$ then $f_n(-1) > 0 > f_n(1)$ , and so one integer root must be $0$ . This implies that $n=0$ . But $f_0(x)$ does not have three integer roots. Thus none of $n=-1,0,1$ work, and hence $n = \pm 2$ . Both $f_2(x)$ and $f_{-2}(x)$ have integer roots ( $-2,1,1$ and $2,-1,-1$ respectively), so the answer is $2 \times -2 = \boxed{-4}$ .

Let the roots of equation be

$\alpha, \beta , \gamma$ .

Then we have

$\alpha+ \beta + \gamma= 0.$ .

$\alpha \beta + \beta \gamma + \gamma \alpha= -3$ .

Simplifying above 2 equations one gets

$\alpha^{2} = 3 + \beta \gamma$ , without the loss of generality.

Then we have values of

$\beta \gamma= {-2,1,6,13....}$ .

While $\alpha \in {1,-1,2,-2,3,-3....}$ .

Now $\alpha \beta \gamma$ = ${-2,2,-6,6...}$ ..... $eq^{n}_1$ Now for all solutions to be real we have $\triangle \geq 0$ (cubic discriminant $>0$ ).

From here we find $n \in [-2,2]$ ..... $eq^{n}_2$ Thus $\alpha \beta \gamma$ also lies in same interval . As

$\alpha \beta \gamma$ = $-n$

Thus we get only two values $2,-2$ for $n$ .

Hence answer is $\boxed{-4}$

Note : $\triangle = b^{2}c^{2} -4ac^{3} -4bd^{3} -27a^{2}d^{2}+ 18abcd$