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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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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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