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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?

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For starters $15w^2-14$ is a square for $w=\pm1,\pm3,\pm5$ so you get six by taking $w=2x+1$. But eventually you get to surfaces of general type. See my answer to . – Noam D. Elkies Feb 11 '12 at 23:01
...and if I did it right, you can get to $w = \pm 7$ using points on an elliptic curve of (conductor 360 and) rank 1. That might be the maximum. FWIW $0,\pm1,\pm2,\pm3$ seems to give a curve of conductor 30 with 12 torsion points but rank zero. – Noam D. Elkies Feb 11 '12 at 23:12
up vote 12 down vote accepted

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.

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A recent paper of Gonzales-Jimenez and Xarles [Acta Arith. 149 (2011)] (apologies for the lack of accents) gets a (sharp) upper bound of $8$ for those quadratics with an axis of symmetry half-way between two integers. – Mike Bennett Feb 12 '12 at 4:42
@Mike: That's great, thank you! – GH from MO Feb 12 '12 at 7:19

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