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 May 14th, 2019, 05:55 AM #1 Newbie   Joined: May 2019 From: New York Posts: 1 Thanks: 0 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. Tags common, divisor, equality, greatest, numbers, search, sequences Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Algebra 5 February 14th, 2014 10:15 AM michael1270 Number Theory 0 September 11th, 2013 09:58 AM matqkks Number Theory 2 June 29th, 2013 08:35 AM remeday86 Number Theory 1 June 29th, 2010 08:10 PM remeday86 Applied Math 1 December 31st, 1969 04:00 PM

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