Timeline for Diophantine approximation by fractions whose numerator and denominator are both prime
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4 at 1:57 | answer | added | Matthew Bolan | timeline score: 3 | |
| Sep 3 at 14:42 | comment | added | Stefan Kohl♦ | @MatthewBolan Thank you very much! - Wouldn't you like to turn this into an answer? Besides, I also suspected the third question may be the hardest. | |
| Sep 3 at 13:09 | comment | added | Matthew Bolan | $S$ is dense in $\mathbb R^+$. The point is that $$S_n=\bigcup_{p,q \in P; q > n} \left(\frac{p}{q}-\frac{1}{q^2},\frac{p}{q}+\frac{1}{q^2}\right)\setminus\{\frac{p}{q}\}$$ is open and dense for each $n$, since for large enough $q$ there is always a prime in $(aq,bq)$ for any fixed $b>a\ge 0$. Thus $S = \bigcap_n S_n$ is dense by the Baire Category theorem. I think the third question is potentially hard, but heuristically I don't think any algebraic numbers should work. | |
| Sep 3 at 3:12 | comment | added | Matthew Bolan | Actually, the fact $p,q$ are odd isn't needed to get $|q^2 + pq - q^2| = 1$, even with integer $p,q$ this form does not represent $\pm 2$. | |
| Sep 3 at 3:11 | comment | added | Matthew Bolan | Mathworker21's idea gives an easy proof for $\varphi = \frac{1+\sqrt 5}{2}$. We have $|\varphi - \frac{p}{q}| < \frac{1}{q^2}$ iff $|q^2+pq-p^2| < |\frac{p}{q}-\bar{\varphi}|$. Now when $q$ is large the fact the lefthand side is an odd integer implies $|q^2+pq-p^2| = 1$, so we must in fact satisfy the stronger $|\varphi - \frac{p}{q}| < \frac{1}{2q^2}$. By a theorem of Legendre's, $p/q$ is then a convergent of the continued fraction for $\varphi$, so $q,p$ are consecutive prime Fibonacci numbers. This only happens twice as $F_{n}|F_{2n}$ means $F_{2n}$ is not prime for $n>2$. | |
| Sep 2 at 20:38 | comment | added | Stefan Kohl♦ | @mathworker21 For $x = \sqrt{2}$, one finds the solutions $(p,q) = (3,2)$, $(p,q) = (7,5)$, $(p,q) = (41,29)$, $(p,q) = (63018038201,44560482149)$ and $(p,q) = (19175002942688032928599,13558774610046711780701)$, and there is a chance that these are all (if there are more, $p$ and $q$ have more than $1000$ decimal digits). | |
| Sep 2 at 13:50 | comment | added | mathworker21 | About $x = \sqrt{2}$: one has $|\sqrt{2}-\frac{p}{q}| < \frac{1}{q^2}$ if and only if $|2q^2-p^2| < 2\sqrt{2}+(\frac{p}{q}-\sqrt{2})$, so for large $p,q$, the condition becomes $2q^2-p^2 \in \{-2,-1,0,1,2\}$, but $2q^2-p^2$ will be odd, so either $2q^2-p^2 = -1$ or $2q^2-p^2 = 1$. I think $2q^2-p^2 = -1$ can be ruled out: $2q^2 = (p-1)(p+1)$ implies $q \mid p-1$ or $q \mid p+1$, but $p \sim \sqrt{2}q$, so this is impossible. Maybe $2q^2-p^2 = 1$ can be ruled out by working over $\mathbb{Z}[i]$? | |
| Sep 1 at 22:14 | comment | added | Matthew Bolan | I think that $S$ has Lebesgue measure 0. The numbers in $(-m, m)$ well approximated with denominator a fixed prime $q$ and numerator some prime have measure $ 2m\frac{1+o(1)}{q\log(qm)}$ by PNT. Since $\sum_{q \in P} \frac{1}{q\log(q)} < \infty$ by PNT, Borel-Cantelli finishes. | |
| Sep 1 at 21:13 | comment | added | Stefan Kohl♦ | Numerical experimentation suggests that e.g. for the square root of 2, there is only a small finite number of pairs of primes which work, which is supported by heuristics. | |
| Sep 1 at 12:24 | history | asked | Stefan Kohl♦ | CC BY-SA 4.0 |