Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:45:53Z http://mathoverflow.net/feeds/question/65059 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65059/does-the-quadratic-form-x2-7y2-represent-infinitely-many-primes-with-the-r Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? Frank Thorne 2011-05-15T19:27:53Z 2011-05-16T14:55:28Z <p>Surely yes, and in more generality, but can it be proved?</p> <p>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 $0 &lt; y &lt; x/10$.</p> <p>Some related references which didn't lead to a proof: First of all there is <a href="http://mathoverflow.net/questions/55384/primes-represented-by-two-variable-quadratic-polynomials" rel="nofollow">this</a> previous MO post, which suggests a negative answer.</p> <p>There is also <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa24/aa2451.pdf" rel="nofollow">this paper</a> of Iwaniec, which uses sieve methods but which also uses the multiplicative structure of solutions to the quadratic form.</p> <p>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.</p> <p>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.</p> <p>I wonder if there is a promising approach out there which I have overlooked? Thank you!</p> http://mathoverflow.net/questions/65059/does-the-quadratic-form-x2-7y2-represent-infinitely-many-primes-with-the-r/65074#65074 Answer by Daniel Litt for Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? Daniel Litt 2011-05-15T21:00:59Z 2011-05-15T21:00:59Z <p>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.</p> <p>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. </p> <p>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 $$x'/y'=\frac{8c+21}{3c+8}.$$</p> <p>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 <code>$y&lt;x/(\sqrt{7}-\epsilon)$</code> for any $\epsilon$, unless I've screwed up somewhere.</p> <p>Not as good as $1/10$, but not terrible I guess.</p> http://mathoverflow.net/questions/65059/does-the-quadratic-form-x2-7y2-represent-infinitely-many-primes-with-the-r/65105#65105 Answer by anonymous for Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? anonymous 2011-05-16T03:34:53Z 2011-05-16T03:34:53Z <p>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. </p> http://mathoverflow.net/questions/65059/does-the-quadratic-form-x2-7y2-represent-infinitely-many-primes-with-the-r/65140#65140 Answer by François Brunault for Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? François Brunault 2011-05-16T14:13:14Z 2011-05-16T14:55:28Z <p>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, <em>Algebraic number theory</em>, 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 $\pi$).</p> <p>You want $0 &lt; y &lt; x/10$ which translates into the inequality</p> <p>\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 $\mathbf{R}/\alpha \mathbf{Z}$.</p> <p>Using the equidistribution result above, the set <code>$X=\{\pi : \lambda(\pi) \in [0,\frac12 \log C]\}$</code> 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 <code>$Y=\{\frac12 \log p : \pi \in X\}$</code> 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.</p>