# (Quasi)-Galois radii of an integer

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 $$n$$?

• What have you tried? Starting with $k=2$ seems natural.
– lulu
Jan 20 at 20:46
• 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 $16$. Jan 20 at 20:54
• 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.
– lulu
Jan 20 at 20:59

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.