NEW CONJECTURE: There is no general upper bound.
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.
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, 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
$866988574 = 2 \cdot 433494287.$
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 sequence
http://www.oeis.org/A000229
some 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.
Best Answer
I just wanted to remark that if $p$ is a prime such that $\ell$ splits in $F = \mathbb{Q}(\sqrt{-p})$ for all $\ell \le N$, then one may prove the existence of $N$ consecutive integers which are norms of integers in $\mathcal{O}_F$, providing one is willing to assume a standard hypothesis about prime numbers, namely, Schinzel's Hypothesis H.
First, note the following:
Lemma 1: If $C$ is an abelian group of odd order, then there exists a finite (ordered) set $S = \{c_i\}$ of elements of $C$ such that every element in $C$ can be written in the form $\displaystyle{\sum \epsilon_i \cdot c_i}$ where $\epsilon_i = \pm 1$.
Proof: If $C = A \oplus B$, take $S_C = S_A \cup S_B$. If $C = \mathbb{Z}/n \mathbb{Z}$ then take $S = \{1,1,1,\ldots,1\}$ with $|S| = 2n$.
Let $C$ be the class group of $F$. It has odd order, because $2$ splits in $F$ and thus $\Delta_F = -p$. Let $S$ be a set as in the lemma. Let $A$ denote an ordered set of distinct primes $\{p_i\}$ which split in $\mathcal{O}_F$ such that one can write $p_i = \mathfrak{p}_i \mathfrak{p}'_i$ with $[\mathfrak{p}_i] = c_i \in C$, where $c_i$ denotes a set of elements whose existence was shown in Lemma 1.
Lemma 2: If $n$ is the norm of some ideal $\mathfrak{n} \in \mathcal{O}_F$, and $n$ is not divisible by any prime $p_i$ in $A$, then $$n \cdot \prod_{A} p_i$$ is the norm of an algebraic integer in $\mathcal{O}_F$.
Proof: We may choose $\epsilon_i = \pm 1$ such that $\displaystyle{[\mathfrak{n}] + \sum \epsilon_i \cdot c_i = 0 \in C}$.
By assumption, $[\mathfrak{p}_i] = c_i \in C$ and thus $[\mathfrak{p}'_i] = -c_i \in C$. Hence the ideal $$\mathfrak{n} \prod_{\epsilon_i = 1} { \mathfrak{p}} \prod_{\epsilon_i = -1} \mathfrak{p}'$$ is principal, and has the desired norm.
By the Chebotarev density theorem (applied to the Hilbert class field of $F$), there exists a set $A$ of primes as above which avoids any fixed finite set of primes. In particular, we may find $N$ such sets which are pairwise distinct and which contain no primes $\le N$. Denote these sets by $A_1, \ldots, A_N$.
By the Chinese remainder theorem, the set of integers $m$ such that $$m \equiv 0 \mod p \cdot (N!)^2$$ $$m + j \equiv 0 \mod \prod_{p_i \in A_j} p_i, \qquad 1 \le j \le N$$ is of the form $m = d M + k$ where $0 \le k < M$, $d$ is arbitrary, and $M$ is the product of the moduli.
Lemma 3: Assuming Schinzel's Hypothesis H, there exists infinitely many integers $d$ such that $$ P_{dj}:= \frac{dM + k + j}{j \cdot \prod_{p_i \in A_j} p_i}$$ are simultaneouly prime for all $j = 1,\ldots,N$.
Proof: By construction, all these numbers are coprime to $M$ (easy check). Hence, as $d$ varies, the greatest common divisor of the product of these numbers is $1$, so Schinzel's Hypothesis H applies.
Let $\chi$ denote the quadratic character of $F$. Note that $dM + k + j = j \mod p$, and so $\chi(dM + k + j) = \chi(j) = 1$ (as all primes less than $N$ split in $F$). Moreover, $\chi(p_i) = 1$ for all primes $p_i$ in $A_j$ by construction. Hence $\chi(P_{dj}) = 1$. In particular, if $P_{dj}$ is prime, then $P_{dj}$ and $j \cdot P_{dj}$ are norms of (not necessarily principal) ideals in the ring of integers of $F$. By Lemma 2, this implies that $$dM + k + j = j \cdot P_{dj} \prod_{p_i \in A_j} p_i$$ is the norm of some element of $\mathcal{O}_F$ for all $j = 1,\ldots, N$.
One reason to think that current sieving technology will not be sufficient to answer this problem is the following: when Sieving produces a non-trivial lower bound, it usually produces a pretty good lower bound. However, there are no good (lower) bounds known for the following problem: count the number of integers $n$ such that $n$, $n+1$, and $n+2$ are all sums of two squares. Even for the problem of estimating the number of $n$ such that $n$ and $n+1$ are both sums of squares is tricky - Hooley implies that the natural sieve does not give lower bounds (for reasons analogous to the parity problem). Instead, he relates the problem to sums of the form $\displaystyle{\sum_{n < x} a_n a_{n+1}}$ where $\sum a_n q^n = \theta^2$ is a modular form. In particular, he implicitly uses automorphic methods which won't work with three or more terms.