Сonditions for the equality of Greatest Common Divisor of two specified sequences of numbers

Question

I consider two sequences of numbers $A=\{a_1,...,a_n\}$ and $B=\{k-a_1,...,k-a_n\}$, where $a_1 \le a_2 \le ... \le a_n \le k$.

I am looking for such conditions under which: $gcd(a_1,...,a_n)=gcd(k-a_1,...,k-a_n)=1$.

In more general form: $gcd(a_1,...,a_n)=gcd(k-a_1,...,k-a_n) \ge 1$.

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I found only three particular solutions.

1. If there is such a number $\exists a_s \in A: k-a_t=a_s$, where $a_t \in A$ then $gcd(a_1,...,a_n)=gcd(k-a_1,...,k-a_n)$.

2. Let $gcd(k-a_1,...,k-a_n) = 1$ and $a_i|k, \forall a_i \in A$, then $gcd(a_1,...,a_n) = 1$.

3. Let $gcd(a_1,...,a_n) = 1$ and $k=a_n+1$, then $gcd(k-a_1,...,k-a_n) = 1$.

I am convinced that there are other solutions, but I can not find them yet. I will be grateful for any help.

Answer 1

Here are more particular solutions:

Let $d=\gcd(a_1,\ldots, a_n)$ and $D=\gcd(a_1-a_n,\ldots, a_{n-1}-a_n)$. Let $e(k)=\gcd(k-a_1,\ldots, k-a_n)$.

If $e(k)=d$, then necessarily $d\mid k$. On the other hand, if $d\mid k$, then $d\mid e(k)$, but also $e(k)\mid (k-a_n)-(k-a_i)$ for all $i$ and so $e(k)\mid D$. Hence the following special case:

If $d=D$, then $k$ will do iff $k$ is a multiple of $d$.

Now suppose $d\ne D$, let $p$ be a prime dividing $\frac Dd$, and suppose $pd\mid k$. Then for some $i$, $pd\nmid a_i$, hence $pd\nmid k-a_i$, $pd\nmid e(k)$. If follows that we can make $e(k)=d$ by ensuring $pd\mid k$ for all applicable $p$:

Any $k$ that is a multiple of $$d\cdot\prod_{p\mid \frac Dd}p$$ will do.

Answer 2

Here is a particular solution which has some interesting aspects.

Let $a_i=\frac{p_n\#}{p_i}$ and $k=p_n\#$ where $p_n\#$ denotes the primorial equaling the product of the first $n$ prime numbers and $p_i$ is the $i^{th}$ prime number. Since each $a_i$ is deficient in one prime factor, no prime factor divides every element and the greatest common denominator of all elements is $1$.

With respect to any particular element $a_i$, $$k=p_i\cdot a_i;\space k-a_i=(p_i-1)a_i$$ The differences (being multiples of $a_i$) each still have all of the prime factors of $k(=p_n\#)$ except $p_i$. Thus the greatest common denominator of all elements in the sequence of differences is also $1$. Each factor of $(p_i-1)$ that appears is composite (except for $(p_1-1)=1$ and $(p_2-1)=2$) and will have prime factors smaller than itself that are already present in $a_i$.

It is noteworthy that all of the elements of the first sequence taken pairwise have very large common denominators: $\gcd(a_i,a_j)=\frac{p_n\#}{p_ip_j}$. Hence $\frac{p_n\#}{p_{n-1}p_n}=p_{n-2}\#\le \gcd(a_i,a_j)\le \frac{p_n\#}{p_1p_2}=\frac{p_n\#}{6}$.