Consecutive square values of cubic polynomials Let $P(x)$ be a cubic polynomial with integer coefficients. Does there exist a constant $c$ such that at least one of the following values $P(0),P(1),...,P(c)$ is not a square?  
It is known that the number of integral points of $y^2 = P(x)$ is bounded by a constant dependent on the coefficients of $P(x)$ and I am wondering whether in this particular case, it is possible to find a constant independent of the coefficients. Any links to existing literature or maybe an explanation of why such a problem would be impossible to solve with current tools would be of a great help too.
 A: Lower bound for $c$ is 6 $7$.
$$ f(x) = -5096*x^3 + 70161*x^2 - 232960*x + 262144 $$ is square for $ x \in [0,6]$.
This is specialization of the identity which is square for $x \in [0,5]$:
$$ P_5 = \left(2 b_{0} + \frac{75}{8}\right) x^{3} + \left(-15 b_{0} - \frac{4311}{64}\right) x^{2} + \left(\frac{115}{4} b_{0} + 120\right) x + b_{0}^{2}$$
For $0 \le x \le 7$
$g(x)=4*x^3 - 36*x^2 + 80*x + 1$
If I remember correctly, the related question for quadratic is conjecturally bounded by about $8$.
Added
The $P_5$ identity gave infinite family of $8$ solutions arising
from genus $1$ curve (to avoid scaling require the content to be square free).
I believe absolute bound on $c$ follows from 
the conjectures about absolute bound of number of rational
points on curves of genus $ > 1$.
Since there are no four consecutive squares in arithmetic progression, the cubic $f(x)$ is square-free.
This leads to the curve $[f(x)=y^2,f(x+1)=z^2]$ which I expect to be genus $4$
in general (and possibly always).
A: I show here that the existence of such a constant $c$ (and much more) follows from previous conjectures.
Namely, a special case of the Caporaso-Harris-Mazur conjecture asserts that there is a constant $N$ such that: for any squarefree degree-$6$ polynomial $f(x)\in\mathbf{Q}[x]$, there are at most $N$ rational numbers $b$ for which $f(b)$ is a square (in $\mathbf{Q}$).  So if $P(x)$ is a cubic polynomial in $\mathbf{Q}[x]$ for which $P(x)P(x+1)$ is squarefree, then there are at most $N$ rational numbers $b$ for which both $P(b)$ and $P(b+1)$ are squares, whence you can take $N+1$ to be your constant $c$.  If $P(x)$ is squarefree but $P(x)P(x+1)$ is not, then one can use the above argument with $P(x)P(x+r)$ for some $r\in\{2,3,4\}$, as at least one of these will be squarefree.  If $P(x)$ is not squarefree then we can divide it by a square factor in order to reduce to the analogous question for degree-one polynomials, where it's easy to exhibit such a value $c$.  
