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Say $r$ is a Galois radius of an integer $n$ if $\omega(n-r)=\omega(n+r)=1$, where $\omega$ counts the prime factors regardless of multiplicity, and say $s$ is a $k$-quasi-Galois radius of $n$ if $\omega(\vert n-s\vert)\leq k$ and $\omega(n+s)\leq k$.

Is the product of $k$ pairwise distinct Galois radii of $n$ always a $k$-quasi-Galois radius of $n$?

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    $\begingroup$ What have you tried? Starting with $k=2$ seems natural. $\endgroup$
    – lulu
    Jan 20 at 20:46
  • $\begingroup$ I tried a few computations in my head, starting with $n=16$, whose Galois radii are $3, 7, 9, 11, 13$. Any product of $2$ distinct members of this list gives rise to a $2$-quasi-Galois radius of $16$. $\endgroup$ Jan 20 at 20:54
  • $\begingroup$ Well, starting with such a small number is unlikely to get you anywhere. After all, for small $n$, $\omega(n)≤2$. I suggest taking a number with a large $\omega$ value, like $2\times 3\times 5\times 7\times 11$ and looking at $n$ near that. $\endgroup$
    – lulu
    Jan 20 at 20:59

1 Answer 1

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This is not true, even in the case $k =2$. Let $n = 5379.$ First, note that $5379 - 3248 = 2131$ and $5379 + 3248 = 8627$ are both prime, hence $3248$ is a Galois radius of $5379.$

Similarly, note that $5379 - 3250 = 2129$ and $5379+3250 = 8629$ are also both prime, hence $3250$ is a Galois radius of $5379.$

Now, consider $s = 3248 \cdot 3250 = 10556000$. Calculating, we see that $|5379-s| = 10550621,$ which has prime factorization $$10550621 = 61 \times 257 \times 673.$$ So, we have $\omega(|n-s|) = 3,$ and thus $s$ is not a $2$-quasi-Galois radius of $n$, even though $s$ is the product of two distinct Galois radii of $n$.


The way I obtained this counterexample is as follows. First, I noted that if you have two pairs of twin primes $(p, p+2)$ and $(q, q+2)$, then $n = \frac{p+q+2}{2}$ has two distinct Galois radii, given by $n-(p+2)$ and $n-p$. So, this gave me an easy way to find large numbers with (at least) two distinct Galois radii.

Then, following what lulu noted in the comments (that small numbers don't have many prime factors), I chose a few somewhat large twin prime pairs to mess around with, and came across this counterexample pretty quickly.

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