GCD of Two Specified Sequences of Numbers: Conditions for Equality

[Vitaly1]

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 $gcd(a_1,...,a_n) = 1$.

5. 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.

 

[jvicens]

Thank you for sharing your findings and asking for help in finding other solutions. I am always interested in exploring and discovering new patterns and solutions.

After analyzing your proposed solutions, I have found a few more conditions under which $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n) \ge 1$.

6. If $gcd(a_1,...,a_n) = 1$ and $gcd(k-a_1,...,k-a_n) = d$, where $d$ is a divisor of $k$, then $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n) = d$.

7. Let $m$ be the smallest number in the set $A$, then $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n) = gcd(m,k)$.

8. If $a_n = k$, then $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n) = 1$.

9. Let $p$ be a prime number such that $p|k$ and $p$ does not divide any of the numbers in set $A$, then $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n) = 1$.

10. If $a_i = a_j$ for some $i \neq j$, then $gcd(a_1,...,a_n) = gcd(k-a_1,...,k-a_n)$.

I hope these additional conditions will be helpful in your search for solutions. However, I would also like to mention that finding all possible solutions to this problem may be a challenging task, as there are infinite combinations of numbers that can satisfy the given conditions.

Perhaps, instead of looking for specific conditions, we can try to prove that these conditions are the only ones that satisfy the given equation. This would require a deeper understanding of number theory and the properties of greatest common divisors.

I wish you the best of luck in your search and I am looking forward to any further discussions on this topic.