how many consecutive integers $x$ can make $ax^2+bx+c$ square ?

2012-02-11T22:17:47Z

The following problem was raised in a Mathlinks thread:

If $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b^2-4ac\ne 0$, for how many consecutive integers $x$ can $ax^2+bx+c$ ba a perfect square ?

The polynomial $-15x^2+64$ is obviously a square for the five numbers $x=-2,...,2$, but the method used for finding this in the above thread cannot be extended further. Should $5$ really be the best possible answer?

Has this problem been treated somewhere else?

Answer by GH for how many consecutive integers $x$ can make $ax^2+bx+c$ square ?

2012-02-11T23:43:01Z

To resonate with Noam Elkies' comments, it is conjectured that $8$ squares is the maximum for arbitrary $a$, and $4$ squares is the maximum for $a=1$. For $5$ symmetric squares the smallest known leading coefficients are $a=15$ and $a=-20$, while for $5$ increasing squares they are $a=60$ and $a=-56$. It is known that there are infinitely many examples with $5$ or $8$ symmetric squares, or with $6$ increasing squares. It is also known that there is no symmetric sequence of $7$ squares, and only finitely many of $10$ squares up to obvious equivalences (this one follows from Falting's theorem applied to a specific hyperelliptic curve).

Good starters are:

Browkin-Brzeziński: On sequences of squares with constant second differences, Canad. Math. Bull. 49 (2006), 481–491.

Bremner: On square values of quadratics, Acta Arith. 108 (2003), 95–111.

how many consecutive integers $x$ can make $ax^2+bx+c$ square ? - MathOverflow

how many consecutive integers $x$ can make $ax^2+bx+c$ square ?

Answer by GH for how many consecutive integers $x$ can make $ax^2+bx+c$ square ?