show that $b^2-2b-4ac-7$ is a square number 
Let $ \varphi(x)$ be a cubic polynomial with integer coefficients. Given that $ \varphi(x)$ has $3$ distinct real roots $u,v,w $ and that $u,v,w $  are not rational. Suppose there are integers  $ a, b,c$  such that $u=av^2+bv+c$.  Prove that $b^2 -2b -4ac - 7$ is a square number .

I only know  $\varphi(x)$ is the minimal polynomial of its roots over $\mathbb Q$ (down to monicity of course).   But then I don't have much idea, please help.
 A: Apologies for the following, which is a little messy.
We may consider the slightly more general case when $a$, $b$, and $c$ are all rational numbers.
Let $\Delta = b^2 - 2b - 4ac - 7.$
We make the stronger claim, namely, that if
$$T = u + v + w \in \mathbf{Q},$$
then $\Delta = \Phi^2$, where
$$\Phi = 3b + 1 + 2 a T.$$
The irreducibility assumption implies that $K = \mathbf{Q}(v)$ has degree three. The fact that $u = a v^2 + b v + c \in K$ implies that $K$ is Galois with Galois group cyclic of order $3$. Moreover, a generator of the Galois group is given by $\phi(v) = a v^2 + b v + c$.
Suppose one replaced $u$ by $u - \lambda$ for a rational scalar $\lambda$. This replaces
$$(a,b,c,T) \mapsto (a,b + 2 a \lambda,c - \lambda+ b \lambda + a\lambda^2,T - 3 \lambda).$$
One can easily check that $\Delta$ and $\Phi$ remain unchanged. Similarly, if one replaces $u$ by $u \lambda$, then
$$(a,b,c,T) \mapsto (a \lambda^{-1},b,c \lambda,T \lambda),$$
and $\Delta$ and $\Phi$ are once more unchanged. Let's make the reduction that $T = 0$, but wait until later to specialize $a$ to $1$.
We have
$$-  a v^2 - (b+1) v  - c = - v - u = w.$$
In particular, this gives an extra symmetry, and shows that our problem is now symmetric in the transformation $(a,b,c) \mapsto (-a,-b-1,-c)$.
Since $w = \phi(u) = \phi^2(v)$, we also have
$$- a v^2 - (b+1) v - c = w = \phi^2(v) 
= a(a v^2 + b v + c)^2 + b (a v^2 + b v + c) + c.$$
This is a quartic in $v$ with coefficients over $\mathbf{Q}$ which we may call $F_{a,b,c}(x)$. Note that this quartic must be divisible by the minimal polynomial of $v$. But given the symmetry we noted above, so must the polynomial $F_{-a,-b-1,-c}(x)$. And we find that:
$$F_{a,b,c}(x) + F_{-a,-b-1,-c}(x) = 2 a^2(x^3 + 0 \cdot x^2 + \ldots )$$
is some explicit cubic with coefficients in $\mathbf{Q}[a,a^{-1},b,c]$ and vanishing $x^2$ coefficient (corresponding to $T = 0$). This must be the minimal polynomial of $v$. At this point, we  are pretty close, but some slightly irritating computations remain. Now we use the above reductions to reduce to the case $a = 1$. Explicitly, we have:
$$F_{1,b,c}(x) + F_{-1,-b-1,-c}(x) = x^3 + (c - 1 - b - b^2) x - c(1 + 2b)/2.$$
We now plug $v^2 + b v + c$ into this polynomial, and then divide by the minimal polynomial to to arrive at
$$\left(\frac{\Delta - \Phi^2}{16}\right)((3 c - 2 b c - 8 b^2 c - 4 c^2)
- (4 b - 6 b^2 - 8 b^3 - 2 c - 4 b c) v - (2 b + 8 b^2 + 4 c) v^2) = 0.$$
Since $1$, $v$, and $v^2$ are linearly independent, either the scalar multiple term is zero (which is what we want), or all the coefficients of the quadratic are zero. It's easy enough to see this can't happen. For example, the $v^2$ coefficient implies that $4c = - 2 b -8 b^2$. Substituting this into the $v$ coefficient leads to $b = 0$, and so $c = 0$. But when $c = b= 0$, the cubic above is $x^3 - x$ which is not irreducible. Hence we are done.
