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

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

2011-05-15T19:27:53Z

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!

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

2011-05-15T21:00:59Z

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.

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

2011-05-16T03:34:53Z

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.

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

2011-05-16T14:13:14Z

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.