$f(x)=x^{3}-3 x+a$. Given that it has $3$ integer roots $\alpha, \beta, \gamma$. Find all possible values of $a$. So, I tried vieta and with some algebra, I got to the point $0=\alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \gamma+\gamma \beta+\alpha \beta)$ where $(\alpha \gamma+\gamma \beta+\alpha \beta)=a$. I don't know how to progress from here
 A: We can solve this without breaking symmetry,
$$\alpha+\beta+\gamma=0$$ squaring this we have
$$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma)=0$$
So
$$\alpha^2+\beta^2+\gamma^2=6$$
now since $\alpha,\beta, \gamma$ are integers we have
the only possible values of $\alpha^2,\beta^2,\gamma^2$ are $0, 1, 4$. And it is easy to see that we must have, up to a permutation,
$$\alpha^2=\beta^2=1, \gamma^2=4$$ and thus
$$\alpha^2\beta^2\gamma^2=4$$  so
$$a=\sqrt{4}=\pm 2$$
A: $$f(x) = (x - \alpha)(x - \beta)(x - \gamma) = x^3 - 3x + a$$
$$f(x) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma = x^3 - 3x + a$$
Matching up the coefficients gives you the system of equations:
$$\alpha + \beta + \gamma = 0$$
$$\alpha\beta + \alpha\gamma + \beta\gamma = -3$$
$$\alpha\beta\gamma = -a$$
From the first one, we get $\gamma = -(\alpha + \beta)$.  Substitute into the other two.
$$\alpha\beta - (\alpha + \beta)(\alpha + \beta) = -3 \implies - \alpha^2 - \alpha\beta - \beta^2 = -3 \implies \alpha^2 + \beta^2 + \alpha\beta = 3$$
$$\alpha\beta(-(\alpha + \beta)) = -a \implies \alpha^2\beta + \alpha\beta^2 = a$$
Expressing the first one of these as a quadratic in terms of $\beta$, we get:
$$\beta^2 + \alpha\beta + (\alpha^2 - 3) = 0 \implies \beta = \frac{-\alpha \pm \sqrt{\alpha^2 - 4(\alpha^2 - 3)}}{2} = \frac{-\alpha \pm \sqrt{12 - 3\alpha^2}}{2}$$
This means that in order for $\beta$ to be a real number, we must have $\alpha^2 \le 4$, or $|\alpha| \le 2$.  Since we're given that $\alpha$ is an integer, it follows that $\alpha \in \{-2, -1, 0, 1, 2\}$.  But we also need $\beta$ to be an integer, so $\alpha = 0$ is not valid, as then we'd be dealing with $\sqrt{12} = 2\sqrt{3}$.
Since the equations are symmetric in $\alpha$ and $\beta$, we must also have $\beta \in \{-2, -1, 1, 2\}$.
Now, for each of the possible combinations of ($\alpha$, $\beta$), we can calculate $\gamma = -(\alpha + \beta)$, and then check if the ($\alpha$, $\beta$, $\gamma$) triple follows the constraint $\alpha\beta + \alpha\gamma + \beta\gamma = -3$.  It turns out that there are six valid combinations:

*

*$(-2, 1, 1)$

*$(-1, -1, 2)$

*$(-1, 2, -1)$

*$(1, -2, 1)$

*$(1, 1, -2)$

*$(2, -1, -1)$
From each one, we can calculate $a = -\alpha\beta\gamma$.  And it turns out that there are only two possibilities, $a = \pm 2$.
A: From the missing equation let set $\gamma=-S$ where $\begin{cases}S=\alpha+\beta\\P=\alpha\beta\end{cases}\ $ so as to work with sum and product.
The equation can be rewritten: $$x^3-(S^2-P)x-SP=x^3-3x+a$$
So we continue by replacing $P=S^2-3$ to get $a=-SP=S(3-S^2)$
The new factorization becomes $$x^3-3x+a=(x-S)(x^2+Sx+S^2-3)=0$$
To have $3$ integer solutions we need the discriminant of the quadratic polynomial to be a perfect square:
$\Delta=S^2-4(S^2-3)=12-3S^2$  so $0\le |S|\le 2$ and only $S=\pm 1$ and $S=\pm 2$ work.
Verification:

*

*$S\in\{1,-2\}$ then $a=S(3-S^2)=2$ and $x^3-3x+2=(x+2)(x-1)^2$

*$S\in\{-1,2\}$ then $a=S(3-S^2)=-2$ and $x^3-3x-2=(x-2)(x+1)^2$
A: We can also work with the properties of the function itself.  For $ \ \phi(x) \ = \ x^3 - 3x \ \ , $ the three zeroes are $ \ 0 \ , \ \pm \sqrt3 \ \ ,  $ so the $ \ a \ = \ 0 \ $ case for $ \ f(x) \ $ does not have only integral zeroes.  The turning points of the function curve are easy to find using calculus, but they can also be found through the use of inequalities.
I will give the proof of a proposition (from Korovkin's little book, Inequalities), since it may be unfamiliar.
Theorem:  For $ \ x \ \ge \ 0 \ , \ k \ > \ 0 \ , \ n \ > \ 1 \ , \ $ the function $ \ g(x) \ = \ x^n - kx \ $ has its relative minimum $ \ (1 - k)·\left(\frac{k}{n} \right)^{n/(n-1)} \ $ at $ \ x \ = \ \left(\frac{k}{n} \right)^{1/(n-1)} \ \ . $
We start from the better-known Bernoulli inequality $ \ (1 + z)^p \ \ge \ 1 + pz \ , \ p \ > \ 1 \ , \ z \ \ge \ -1 \ \ . $  We substitute $ \ y \ = \ 1 + z \ $ to write
$$   y^p \ \ \ge \ \  1 + p·(y - 1) \ \ \Rightarrow \ \ y^p \ - \ py \ \ \ge \ \ 1 - p \ \ , \ \ y \ \ge \ 0 \ \ . $$
Multiplying through by $ \ c^p \ , \ c \ > \ 0 \ , $ we have
$$   (cy)^p  \ - \ p·c^{p-1}·(cy) \ \ \ge \ \ (1 - p)·c^p \ \ , \ \ y \ \ge \ 0 \ \ . $$
Finally, taking $ \ x \ = \ cy \ $ and $ \ k \ = \ p·c^{p-1} \ \Rightarrow \ c \ = \ \left(\frac{k}{p} \right)^{1/(p - 1)} \ \ , \ $  we obtain
$$   x^p  \ - \ k·x \ \ \ge \ \ (1 - p)·\left[ \ \left(\frac{k}{p} \right)^{1/(p - 1)} \ \right]^p \ \ = \ \  (1 - p)· \left(\frac{k}{p} \right)^{p/(p - 1)}  \ \ , \ \ x \ \ge \ 0 \ \ , $$
with equality holding for $  \ x \ = \ c \ = \ \left(\frac{k}{p} \right)^{1/(p - 1)} \ \ . \ \Box $
[Of course, differential calculus quickly locates the turning points of $ \ g(x) \ $ from $ \ g'(x) \ = \ n·x^{n-1} - k \ = \ 0 $ $ \Rightarrow \ x \ = \ (\pm) \ \left(\frac{k}{n} \right)^{1/(n-1)} \ \ , $ the plus-or-minus being applicable if $ \ n \ $ is an odd integer.]
Consequently, the relative minimum ("lower turning point") of $ \ \phi(x) \ = \ x^3 \ -  \ 3x \ $ is found at $ \ \left(\left(\frac33 \right)^{1/2} \ , \ (1-3)·\left(\frac33 \right)^{3/2} \right) \ = \ (1 \ , \ -2) \ \ .  $  Because this function has odd symmetry, it has its relative maximum ("upper turning point") at $ \ (-1 \ , \ 2) \ \ . $
As to the function $ \ f(x) \ = \ x^3 - 3x + a \ \ , $ the requirement that there be three integer zeroes places an extremely stringent constraint upon possible values of $ \ a \ \ . $  The Viete relation $ \ \alpha·\beta·\gamma \ = \ -a \ \ $ then implies that $ \ a \ $ must be an integer.  "Shifting" the function curve vertically is limited to the extent that one of the turning points becomes an $ \ x-$intercept, so that there is one "double zero" and one other real zero (there is no requirement that the zeroes be distinct).  Hence, we are restricted to $ \ -2 \ \le \ a \ \le \ +2 \ \ . $
We have already seen that $ \ a \ = \ 0 \ $ possesses only one integer zero.  It is easily determined that the $ \ a \ = \ \pm 2 \ $ cases do have three integer zeroes.  For $ \ a \ = \ 2 \ , $ the function curve is shifted so that the lower turning point lies at $ \ (1 \ , \ 0 ) \ : $ this makes $ \ x \ = \ 1 \ $ a "double zero" (which can also be checked by polynomial/synthetic division), giving us the factorization $ \ x^3 - 3x + 2 \ = \ (x - 1)^2·(x + 2) \ \ . $  We also see that $ \ 1·1·(-2) \ = \ -(2) \ \ . $  By the same token, $ \ a \ = \ -2 \ $ places the upper turning point at $ \ (-1 \ , \ 0) \ \ , $ from which we obtain $ \ x^3 - 3x - 2 \ = \ (x + 1)^2·(x - 2) \ \ , $  with the zeroes obeying $ \ (-1)·(-1)·2 \ = \ -(-2) \ \ . $
Any remaining possibilities are then viewed in the light of $ \ a \ $ only being permitted to take on the values $ \ \pm 1 \ $ or that the integer values of the zeroes must lie in the interval $ \ -2 \ < \ \alpha \ , \ \beta \ , \ \gamma \ < \ +2 \ \ .  $  The zeroes must now also (somehow) be distinct.
If we take $ \ a \ = \ 1 \ \ , \ \ x^3 - 3x + 1 \ \ $ has its turning points at $ \ (-1 \ , \ 3) \ $ and $ \ (+1 \ , \ -1 \ ) \ \ ; $ it is also clear that $ \ x \ = \ 0 \  ,  \ \pm 2 \ $ are not zeroes of the polynomial.  A similar argument can be made for $ \ a \ = \ -1  $ $ \rightarrow \ x^3 - 3x - 1 \ \ . \ $  Alternatively, with $ \ a \ \neq \ 0 \ , \ x \ = \ 0 \ $ cannot be a zero.  But with only $ \ x \ = \ +1 \ , \ -1 \ \ $ available as integer zeroes, there is no combination of three distinct values possible. (Or there is no product of three factors $ \ ( x - 1 ) \ $ and $ \ ( x + 1 ) \ $ that equal a polynomial of the form $ \ x^3 - 3x + a \ \ , \ $ or it is not possible to produce a "Viete sum" $ \ \alpha + \beta + \gamma \ = \ 0 \ \ . \ )  $
We conclude that the only permissible values of $ \ a \ $ are $ \ 2 \ $ and $ \ -2 \ \ . $
