5 added 786 characters in body

NEW CONJECTURE: There is no general upper bound.

EDIT, Thursday 17 June. Wadim Zudilin has been running one of my programs on a fast computer. We finally have a string of 11, the form being $3 x^2 + x y + 26 y^2$ of discriminant $-311.$ The integrally represented numbers start at 897105813710 and end at 897105813720. Note that the maximum possible for this discriminant is 11. So we now have this conjecture: For discriminants $\Delta$ with absolute values in this sequencehttp://www.research.att.com/~njas/sequences/A000229some form represents a set of $N$ consecutive integers, where $N$ is the first quadratic nonresidue. As a result, we conjecture that there is no upper bound on the number of consecutive integers that can be represented by a positive quadratic form.

4 string of 10, also title to 14

# Can a positive binary quadratic form represent 1314 consecutive numbers?

Wadim Zudilin suggested that I make this a separate question. This follows http://mathoverflow.net/questions/23690/ 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.

EDIT, Monday 10 May. I had an idea that the large values represented by any individual form ought to be isolated. That was naive. Legendre showed that for a prime $q \equiv 7 \pmod 8$ there exists a solution to $u^2 - q v^2 = 2,$ and therefore infinitely many solutions. This means that the form $x^2 + q y^2$ represents the triple of consecutive numbers $q v^2, 1 + q v^2, u^2$ and then represents $4 + q v^2$ after perhaps skipping $3 + q v^2$. Taking $q = 8 k - 1,$ the form $x^2 + x y + 2 k y^2$ has no restrictions $\pmod 8,$ while an explicit formula shows that it represents every number represented by $x^2 + q y^2.$ Put together, if $8k-1 = q$ is prime, then $x^2 + x y + 2 k y^2$ represents infinitely many triples. If, in addition, $( 3 | q) = 1,$ it seems plausible to expect infinitely many quintuples. It should be admitted that the recipe given seems not to be a particularly good way to jump from length 3 to length 5, although strings of length 5 beginning with some $q t^2$ appear plentiful.

EDIT, Monday Tuesday 11 May. I have found a string of 9, the form is $6 x^2 + x y + 13 y^2$ and the numbers start at $1786879113 = 3 \cdot 173 \cdot 193 \cdot 17839$ and end with $1786879121$ which is prime. As to checking, I have a separate program that shows me the particular $x,y$ for representing a target number by a positive binary form. Then I checked those pairs using my programmable calculator, which has exact arithmetic up to $10^{10}.$

EDIT, Saturday 15 May. I have found a string of 10, the form is $9 x^2 + 5 x y + 14 y^2$ and the numbers start at $866988565 = 5 \cdot 23 \cdot 7539031$ and end with $866988574 = 2 \cdot 433494287.$

3 string of 9

Wadim Zudilin suggested that I make this a separate question. This follows http://mathoverflow.net/questions/23690/ 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.

EDIT, Monday 10 May. I had an idea that the large values represented by any individual form ought to be isolated. That was naive. Legendre showed that for a prime $q \equiv 7 \pmod 8$ there exists a solution to $u^2 - q v^2 = 2,$ and therefore infinitely many solutions. This means that the form $x^2 + q y^2$ represents the triple of consecutive numbers $q v^2, 1 + q v^2, u^2$ and then represents $4 + q v^2$ after perhaps skipping $3 + q v^2$. Taking $q = 8 k - 1,$ the form $x^2 + x y + 2 k y^2$ has no restrictions $\pmod 8,$ while an explicit formula shows that it represents every number represented by $x^2 + q y^2.$ Put together, if $8k-1 = q$ is prime, then $x^2 + x y + 2 k y^2$ represents infinitely many triples. If, in addition, $( 3 | q) = 1,$ it seems plausible to expect infinitely many quintuples. It should be admitted that the recipe given seems not to be a particularly good way to jump from length 3 to length 5, although strings of length 5 beginning with some $q t^2$ appear plentiful.

EDIT, Monday 11 May. I have found a string of 9, the form is $6 x^2 + x y + 13 y^2$ and the numbers start at $1786879113 = 3 \cdot 173 \cdot 193 \cdot 17839$ and end with $1786879121$ which is prime. As to checking, I have a separate program that shows me the particular $x,y$ for representing a target number by a positive binary form. Then I checked those pairs using my programmable calculator, which has exact arithmetic up to $10^{10}.$

2 triples, quintuples
1