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Will Jagy
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Can a positive binary quadratic form represent 13 consecutive numbers?

Wadim Zudilin suggested that I make this a separate question. This follows representability of consecutive integers by a binary quadratic form where most of the people who gave answers are worn out after arguing over indefinite forms and inhomogeneous polynomials. Some real effort went into this, perhaps it will not be seen as a duplicate question.

So the question is, can a positive definite integral binary quadratic form $$ f(x,y) = a x^2 + b x y + c y^2 $$ represent 13 consecutive numbers?

My record so far is 8: the form $$6x^2+5xy+14y^2 $$ represents the 8 consecutive numbers from 716,234 to 716,241. Here we have discriminant $ \Delta = -311,$ and 2,3,5,7 are all residues $\pmod {311}.$ I do not think it remotely coincidental that $$6x^2+xy+13 y^2 $$ represents the 7 consecutive numbers from 716,235 to 716,241.

I have a number of observations. There is a congruence obstacle $\pmod 8$ unless, with $ f(x,y) = a x^2 + b x y + c y^2 $ and $\Delta = b^2 - 4 a c,$ we have $\Delta \equiv 1 \pmod 8,$ or $ | \Delta | \equiv 7 \pmod 8.$ If a prime $p | \Delta,$ then the form is restricted to either all quadratic residues or all nonresidues $ \pmod p$ among numbers not divisible by $p.$

In what could be a red herring, I have been emphasizing $\Delta = -p$ where $p \equiv 7 \pmod 8$ is prime, and where there is a very long string of consecutive quadratic residues $\pmod p.$ Note that this means only a single genus with the same $\Delta = -p,$ and any form is restricted to residues. I did not anticipate that long strings of represented numbers would not start at 1 or any predictable place and would be fairly large. As target numbers grow, the probability of not being represented by any form of the discriminant grows ( if prime $q \parallel n$ with $(-p| q) = -1$), but as the number of prime factors $r$ with $(-p| r) = 1$ grows so does the probability that many forms represent the number if any do. Finally, on the influence of taking another $\Delta$ with even more consecutive residues, the trouble seems to be that the class number grows as well. So everywhere there are trade-offs.

Will Jagy
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