On a problem that equates $\frac{\text{prime}-1}{\operatorname{rad}(\text{prime}-1)}$ with the sequence of primorials

Question

We denote for integers $m>1$ the product of the distinct prime numbers dividing $m$ as $$\operatorname{rad}(m)=\prod_{\substack{p\mid m\\p\text{ prime}}}p,$$ with the definition $\operatorname{rad}(1)=1$ (see it you want the Wikipedia Radical of an integer), that is the famous multiplicative function in the statement of the abc conjecture. Also we denote the $k$th primorial as $$N_k=\prod_{t=1}^k p_t$$ that is the the product of the first $k$ primes.

I was inspired in the first paragraph of section B46 of [1] to propose a variant of the type of problems that was shows the book. I consider solutions $(p,k)$, where $p$ denotes always a prime number for which the identity $$\frac{p-1}{\operatorname{rad}(p-1)}=N_k\tag{1}$$ holds for some primorial $N_k$, with $k\geq 1$.

I've used a Pari/GP program to write the first primes satisfying $(1)$, those first few primes $p$'s are (this is a sample of my computations) $5,13,29,37,53,61,149,157,173,181,229, 269,277,$ $293,317,\ldots$ that correspond to these indexes $k$'s $1,1,1,2,1,1,1,1,1,2,1,1,1,1,1\ldots$ of our primorials $N_k$ satisfying the equation $(1)$.

The curiosity that I've thus is what about the size or cardinality of the set $$\mathcal{K}=\{k\geq 1:(p,k)\text{ is a solution of the equation }(1)\}.$$

Conjecture. The set $\mathcal{K}$ is finite.

What I am saying is that thus the set $\mathcal{K}$ remains as a bounded/finite set of positive integers when the other variable $p$ runs over the set of all prime numbers.

Question. Is it possible to refute previous conjecture? Can you provide some reasoning or calculations to elucidate the veracity of the conjecture? Many thanks.

Maybe it can be useful state bounds or an approximation for the cardinality $|\mathcal{K}|$ for increasing segments of primes $2\leq p\leq X$ and primorials with indexes $1\leq k\leq Y$

In this ocassions I have no a great intuition about if previous conjecture holds, but I've stated it, I just to find solutions forthe integers $k=1,2$ and $3$ (I don't know if my post about this sequence has a good mathematical content, the comments are welcome).

References:

[1] Richard K. Guy, Unsolved Problems in Number Theory, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).

Answer 1

The conjecture is false, and in fact every positive integer $k\geq 1$ lies in $\mathcal{K}$. Here is a proof.

Let us fix $k\geq 1$. Let $\pi(x;N_k^3,N_k^2+1)$ be the number of primes $p\leq x$ such that $p\equiv N_k^2+1\pmod{N_k^3}$. Let $\pi_1(x;N_k^3,N_k^2+1)$ be the same with the restriction that $q^2\nmid p-1$ for any prime $q>p_k$. Following the proof of Proposition 15.12 in the book Koninck-Luca: Analytic number theory - Exploring the anatomy of integers (AMS, 2012), we can prove that $$\frac{\pi_1(x;N_k^3,N_k^2+1)}{\pi(x;N_k^3,N_k^2+1)}\to\prod_{p>p_k}\left(1-\frac{1}{p(p-1)}\right)\qquad\text{as}\qquad x\to\infty.$$ Indeed, we only need to modify this proof at a few places: in the definition of $\mathcal{P}_2(x)$ we restrict the prime $q$ by $p_k<q\leq z$, while in (15.18) we restrict $d>1$ to integers composed of prime factors $p_k<q\leq z$ and replace $\pi(x;d^2,1)$ by $\pi(x;N_k^3d^2,a_d)$, where $a_d\equiv N_k^2+1\pmod{N_k^3}$ and $a_d\equiv 1\pmod{d^2}$. Noting that, by Dirichlet's prime number theorem, $$\frac{\pi(x;N_k^3d^2,a_d)}{\pi(x;N_k^3,N_k^2+1)}\to\frac{\phi(N_k^3)}{\phi(N_k^3d^2)}=\frac{1}{\phi(d^2)}\qquad\text{as}\qquad x\to\infty,$$ the rest of the argument goes almost verbatim the same as in the referenced proof. From this density calculation, and Dirichlet's prime number theorem, it follows that there exists a prime $p$ such that $p\equiv N_k^2+1\pmod{N_k^3}$ and $q^2\nmid p-1$ for any prime $q>p_k$. That is, $p=N_k^2M+1$, where $M\equiv 1\pmod{N_k}$ and $M$ is square-free. However, this implies that $$\frac{p-1}{\operatorname{rad}(p-1)}=\frac{N_k^2M}{N_kM}=N_k,$$ whence $k\in\mathcal{K}$. The proof is complete.