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May 14th, 2019, 05:55 AM   #1
Joined: May 2019
From: New York

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Greatest Common Divisor of two specified sequences of numbers (search for equality)

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

I found only five 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 $e|k$, 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 $a_i|k, \forall a_i \in A$, then $gcd(a_1,...,a_n) = 1$.

4. 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.
VitalyGrigoriev is offline  

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