distance between irreducible elements in a number ring Consider the number ring $\mathbb{Z}[\phi]$ where $\phi$ is the positive root of $X^2-X-1$.
Any of its elements can be written as $a+b\phi$ with $a$ and $b$ integers. There is a norm $N$ such that $N(a+b\phi)=|a^2+ab-b^2|$. The norm is multiplicative and satisfies $N(\phi)=1$.
Let $p$ be a prime number greater than 5, which is a square mod 5. A consequence of Dirichlet's unit theorem is that there are two elements x and y such that the elements whose norm is $p$ are those of the form $\pm x \phi^k$ or $\pm y \phi^k$, with $k$ an integer. For instance, with $p=11$, one may generate the set of elements of norm 11 with $3+\phi$ or $4-\phi$ (we just have to check that these values are not the same up to a power of $\phi$). More generally, it is always possible to find such $x$ and $y$, positives, with the additional property that $xy=p$.
Since from the point of view of the norm, the elements $x$ and $x\phi$ are the same, it is quite natural to consider the elements $\log_{\phi}x \mod 1$ and $\log_{\phi}y \mod 1$ on the unit circle. So here comes my question : is it true that, taking all $p$ primes>5 that are squares mod 5, the set of distance between those values is dense in ]0,1/2[? Thanks by advance for any hint or comment.

Call $g(x)=\log_\phi x$ and $f(x)=g(x)\bmod1$ and let $S=\{n\in\Bbb P:(\frac n5)=1\}$ be the set whose image we want to prove dense. It follows from the density of $f(S)$ on $[0,1]$ that $d(f(S),f(S))$ is dense in $[0,\frac12]$ (just pick one element to be fixed and consider its distance to every other element). The density of $f(S)$ follows from:


*

*$g$ is monotone increasing

*$g$ is unbounded

*Differences between adjacent elements of $g(S)$ tends to zero.


Only the last is in question. Since the primes are equidistributed across different remainder classes (Chebotarev's density theorem), gaps in $S$ carry the same asymptotic properties as the prime gaps themselves. From the prime number theorem, the quotient $g_n/p_n\to0$, where $g_n$ is the $n$-th prime gap; the same can be said about $S$, so $\frac{S_{n+1}-S_n}{S_n}\to0\implies\log_\phi S_{n+1}-\log_\phi S_n\to0$, and hence the difference of log primes goes to 0.
