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If we have a polynomial of degree $2$ with rational coefficients, it may or may not be a square of a degree $1$ polynomial. Say we look at $f(0),f(1),f(2),\ldots$ and observe that they are all rational squares.

How many consecutive values do we have to check in order to deduce that the polynomial is a square (and so, that every value will be a square) ?

The corresponding problem for degree $1$ polynomials is known (you can't have $4$ squares in a row) but is not very easy to show.

Also the answer is obviously dependant on the base field.

Here in our case, a sequence of $n$ squares in a row corresponds to rational points on a surface $S_n$ in $\Bbb P^{n-1}$ which is the intersection of $n-3$ quadratic equations. $S_n$ contains the curve $C_n$ of arithmetic progressions of length $n$, which is the intersection of $S_n$ with one hyperplane (many choices there, if I'm not mistaken).

I believe $S_4$ is equivalent over $\Bbb Q$ to $\Bbb P^2$, and so $S_4(\Bbb Q)$ has strictly more points than $C_4(\Bbb Q)$.
Since I know next to nothing about surfaces, I don't know what to do about finding a $n$ such that $S_n(\Bbb Q) = C_n(\Bbb Q)$.

mercio
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    This earlier question looks relevant: http://math.stackexchange.com/questions/333253/a-quadratic-polynomial-getting-square-values-in-consecutive-points – lulu Dec 07 '15 at 15:14
  • Thanks, at a glance it looks like they restricted the polynomials to symmetric polynomials. Doing so brings down the dimension by $1$ and relates the problem to the study of some curves. So there are sequences of $8$ squares in some quadratic progression. And there is no symmetric sequence of $9$ or more squares. – mercio Dec 07 '15 at 15:21
  • Yes, at least that's what I recall from the paper. Not sure if the methods therein generalize or not. Anyway, might be worth a read. – lulu Dec 07 '15 at 15:27

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