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: $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n) \ge 1$.

It can be seen that the problem is solved only in a particular forms.

I found only four 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 $\gcd(a_1,...,a_n)=e$ and $\gcd(a_n-a_1,...,a_2-a_1)=E$. If $e=E$ and $e\mid k$, then $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n)$.

  3. Let $P=p_1 \cdot ... \cdot p_n$ denotes the primorial equaling the product of the first $n$ prime numbers and $p_i$ is the $i^{th}$ prime number. Let $a_i=\frac{P}{p_i}$ and $k=P$, then $\gcd(a_1,...,a_n) = \gcd(k-a_1,...,k-a_n) = 1$.

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

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

  • 1
    $\begingroup$ What if I took $k=a_n+1$? $\endgroup$ – Mohan May 13 '19 at 22:53
  • $\begingroup$ I agree, also a decision on the similarity of 4. But very trivial. $\endgroup$ – Виталий May 13 '19 at 23:13
  • $\begingroup$ Why have you written $\equiv$ here? Normally I'd think that denoted congruence modulo some number but there's no $\pmod{m}$ here... $\endgroup$ – Daniel McLaury May 13 '19 at 23:33

Here is a try. Call $k$ to be good, if $d_k\triangleq (k-a_1,\dots,k-a_n)>(a_1,\dots,a_n)=1$. If $k$ is good, it then follows that, there is a prime $p$, such that $p\mid k-a_i$ for every $1\leqslant i \leqslant n$. In particular, $a_1\equiv \cdots\equiv a_n\equiv k\pmod{p}$. Hence, if $k$ is good, then there necessarily is a prime $p$ at which $a_1,\dots,a_n$ are all congruent.

Now, if $k$ is such that, if for all prime $p$, there is an $i$ with $k-a_i\not\equiv 0\pmod{p}$, we then have that $(k-a_1,\dots,k-a_n)=1$. In particular, a sufficient condition is as follows. If the collection $(a_1,\dots,a_n)$ is such that, for every prime $p$, there exists $i <j$ such that $a_i\not\equiv a_j\pmod{p}$, we have that for any $k$, $(k-a_1,\dots,k-a_n)=1$.

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  • $\begingroup$ Thank you, I understood the point. But what does the expression "k is good" mean? $\endgroup$ – Виталий May 13 '19 at 23:44
  • $\begingroup$ Just a vocabulary that, if $d_k=(k-a_1,\dots,k-a_n)>1$. Nothing super fancy/necessary either. $\endgroup$ – kawa May 14 '19 at 1:28
  • $\begingroup$ Let $P=\{p_1,...,p_s\}, p_s \le a_1$. If for each $p \in P$ in the set ${a_1, ..., a_n}$ exists at least one pair $a_i, a_j: a_i \not\equiv a_j \pmod{p}$ on specific $p$ then $\gcd\{a_1, ..., a_n\}=1$. Only one thing I don't understand - is how to prove that if $a_i \not\equiv a_j \pmod{p}$ then also for the same $p: k-a_i \not\equiv k-a_j \pmod{p}, \forall k \in N$. $\endgroup$ – Виталий May 14 '19 at 15:05
  • $\begingroup$ Because, if $p\mid k-a_i$ for every $i $, if holds that, $a_1\equiv a_2\equiv \cdots \equiv k\pmod{p}$, right? Hence, as long as you ensure $(a_1,\dots,a_n)$ has two distinct elements modulo $p$ for every $p$ prime, you have that for every $k$, $(k-a_1,\dots,k-a_n)=1$, no? $\endgroup$ – kawa May 14 '19 at 19:37
  • $\begingroup$ This not valid for $n=a_1=2$. For example, $\gcd\{2,9\}=1$ but if $k=16$, then $\gcd\{7,14\}=2$. $\endgroup$ – Виталий May 15 '19 at 15:21

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