## Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$?

Surely yes, and in more generality, but can it be proved?

It seems that most, if not all, statements about quadratic forms representing primes fall back on algebraic number theory (i.e. splitting of primes in $\mathbb{Q}(\sqrt{7})$) for their proofs, and so are incompatible with the condition that $0 < y < x/10$.

Some related references which didn't lead to a proof: First of all there is this previous MO post, which suggests a negative answer.

There is also this paper of Iwaniec, which uses sieve methods but which also uses the multiplicative structure of solutions to the quadratic form.

There is also the interesting Theorem 5.36 of Iwaniec and Kowalski, which states that the arguments of prime elements of $\mathbb{Z}[i]$ are equidistributed in $(0, 2\pi)$. This is proved using the Hecke $L$-function $\sum_{\alpha \in \mathbb{Z}[i]} \big( \frac{\alpha}{|\alpha|} \big)^{ik} |\alpha|^{-s}$, for all $k$ divisible by 4. This generalizes further, but presumably not to real quadratic fields, where the infinite unit group would foul the construction up.

Finally, using a straight-up sieve (with only the additive structure of solutions to $x^2 - 7 y^2$) seems hopeless, as sieves tend to be bad at finding primes. There is the recent work of Friedlander-Iwaniec on $x^2 + y^4$ and Heath-Brown on $x^3 + 2y^3$, but these use algebraic number theory in $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{-3})$, and seem unlikely to generalize here.

I wonder if there is a promising approach out there which I have overlooked? Thank you!

-
Regarding "It seems that most, if not all, statements about quadratic forms representing primes fall back on algebraic number theory (i.e. splitting of primes in Q(7√)) for their proofs, and so are incompatible with the condition and so are incompatible with the condition that 0<y<x/10." This isn't obviously true, for example see "On the relative sizes of A and B in p=A²+B², where p is a prime=1(mod 4)" math.carleton.ca/~williams/papers/pdf/237.pdf I guess Daniel Litt's comment is in similar spirit. – Jonah Sinick May 15 2011 at 21:36
Quibble: Heath-Brown's prime-capturing polynomials is $x^3+2y^3$, not $x^3+y^3$ which is reducible and hence won't take prime values. – Greg Martin May 15 2011 at 22:06
I'm not sure of the details, but a reference for this might be in Lang, Algebraic number theory, Chap. XV, see Example 3 at the end of the chapter, in which equiditribution of Gaussian primes is (in some sense) generalized to the case of number fields with class number 1. – François Brunault May 16 2011 at 13:05
I guess we were not very careful there with precise attributions for fairly classical things, but let's make it clear that Th. 5.36 in my book with Iwaniec is a theorem at least as old as Hecke (very probably due to him, but I haven't actually checked the history.) – Denis Chaperon de Lauzières May 16 2011 at 21:34
@Emmanuel : On the other hand, I'm happy that you and Iwaniec have included this result in your book, because this theorem deserves to be known more widely. I was once interested in the distribution of rational points on the unit circle and I had a hard time finding references, until I found it in Lang... – François Brunault May 17 2011 at 5:37

The units of $k=\mathbf{Q}(\sqrt{7})$ have the form $\pm (8+3 \sqrt{7})^n$ with $n \in \mathbf{Z}$. If $\pi = x+y\sqrt{7}$ is a prime element of $k$, then $\lambda(\pi):= \log |x+y\sqrt{7}|$ is well-defined in $\mathbf{R}/\alpha \mathbf{Z}$ where $\alpha = \log(8+3\sqrt{7})$. Note that $\lambda$ factors as $\lambda = f \circ \sigma$ where $\sigma : k^{\times} \to \mathbf{R}^{\times}$ is a given embedding of $k$ and $f : \mathbf{R}^{\times} \to \mathbf{R}/\alpha \mathbf{Z}$ is a continuous group homomorphism. We can apply Hecke's theory of equidistribution (see Lang, Algebraic number theory, Chap. XV, especially Example 3 at the end of the chapter) to show that the sequence $\lambda(\pi)$ is equidistributed in $\mathbf{R}/\alpha \mathbf{Z}$ where $\pi$ runs through the primes of $k$ (with respect to the usual ordering on the norm of $\pi$).

You want $0 < y < x/10$ which translates into the inequality

\begin{equation*} \sqrt{p} \leq x+y\sqrt{7} \leq C \sqrt{p} \end{equation*} where $C=\frac{10+\sqrt{7}}{\sqrt{93}}>1$ and $p=N_{k/\mathbf{Q}}(x+y\sqrt{7})$. This in turn is equivalent to $\lambda(\pi) \in [\frac12 \log p , \frac12 \log p + \log C]$ inside $\mathbf{R}/\alpha \mathbf{Z}$.

Using the equidistribution result above, the set $X=\{\pi : \lambda(\pi) \in [0,\frac12 \log C]\}$ has a positive natural density (here we consider only primes of $k$ which don't belong to $\mathbf{Q}$, but this is ok because the norm of a rational prime $p$ is equal to $p^2$, so these rational primes are negligible). Moreover, the set $Y=\{\frac12 \log p : \pi \in X\}$ is dense in $\mathbf{R}/\alpha \mathbf{Z}$ because of the prime number theorem. So we can find infinitely many primes $\pi \in X$ with $\frac12 \log p \in [-\frac12 \log C,0]$ inside $\mathbf{R}/\alpha \mathbf{Z}$, which implies what you want using the above discussion.

-
I hope someone will triple-check this, because somehow it feels too good to be true, but this is stupendous! Thank you. – Frank Thorne May 16 2011 at 20:28
One thing from Lang's book that might be added: The Lang/Hecke argument relies on a Hecke $L$-function, and the Hecke character (where $\pi$ is a prime of $k = \mathbb{Q}(\sqrt{7})$) is given by $\chi(\pi) = |\pi|^{2 \pi i/\log \alpha}$. This is rigged to be trivial on units, hence it gives a character on ideals or on the idele class group. Now $L(s, \chi)$ is a nontrivial Hecke $L$-function, hence holomorphic at $s = 1$, and hence you get equidistribution. – Frank Thorne May 16 2011 at 20:34
@Frank : Yes, of course I should have said that the real heart of the proof lies in the study of the Hecke L-function. – François Brunault May 17 2011 at 5:45
@Frank: I think you need the holomorphicity of all $L(s,\chi^k)$ with $k\in\mathbb{Z}$ at $s=1$. – GH May 18 2011 at 16:28
@GH: Yes indeed. – Frank Thorne May 19 2011 at 16:48

It is true, with the same proof as Iwaniec-Kowalski. Real or complex, the generators of principal primes are equidistributed modulo units. There just happen to be no units in the complex case.

-
I don't see how to write down a convergent Hecke $L$-series which corresponds to that in I-K. Would you please elaborate? – Frank Thorne May 16 2011 at 3:46
@anonymous: I don't buy this. Hecke characters in the real quadratic case don't seem to be any help; "equidistribution modulo units" becomes, after rescaling, equidistribution of (the orbits under the action of the unit group of) the representations of a prime p on a hyperboloid of the form $x^2-dy^2=1$. – David Hansen May 16 2011 at 7:07
I would also be interested in seeing this spelled out. – David Speyer May 16 2011 at 15:19
Actually, I believe this is correct. See François's answer as well as my comment to it. – Frank Thorne May 16 2011 at 20:34
That's what I get for posting comments at three in the morning. :) – David Hansen May 17 2011 at 6:25

This is more of a comment than an answer, but is too long to fit in the comment box. I'll try to replace the $10$ in the statement by $\sqrt{7}-\epsilon$; the strategy is to adjust solutions by units.

Let $K=\mathbb{Q}(\sqrt{7})$ and let $N: K\to \mathbb{Q}$ be the norm map, which sends $x+\sqrt{7}y$ to $x^2-7y^2$ for $x, y\in \mathbb{Q}$. $K$ is a real quadratic field, so the unit group of $K$ has rank $1$, with $8\pm 3\sqrt{7}$ a unit of infinite order.

Now let $p=x^2-7y^2=N(x+\sqrt{7}y)$ with $x, y\in \mathbb{Z}$. Obviously $N((8\pm 3\sqrt{7})^n(x+\sqrt{7}y))=p$ for any $n$, as $8\pm 3\sqrt{7}$ is a unit Setting $x=cy$ for rational $c$ we have that $(8+3\sqrt{7})(x+y\sqrt{7})=x'+y'\sqrt{7}$ satisfies $$x'/y'=\frac{8c+21}{3c+8}.$$

This is a hyperbolic fractional linear transformation with fixed points $\pm \sqrt{7}$ (one of which is repelling, and the other attracting). So this seems to show that simply by adjusting by a unit, one may find $(x, y)$ with $x^2-7y^2=p$ and $y<x/(\sqrt{7}-\epsilon)$ for any $\epsilon$, unless I've screwed up somewhere.

Not as good as $1/10$, but not terrible I guess.

-
When $x,y$ are large with $x^2-7y^2=p$, then $x/y$ is close to $\sqrt 7$, obviously, so I don't think you can answer his question just by adjusting by units. The question is really whether for infinitely many primes $p$ (not all, by any means), the small solutions to $x^2-7y^2=p$ have $x/y$ far away from $\sqrt 7$. It's not as easy at it looks. – Felipe Voloch May 15 2011 at 22:09
I agree that there's no way this method will give $10$ in the denominator; I just figured it was worth observing that one could say something without too much trouble. – Daniel Litt May 15 2011 at 22:15