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#1 h May 29, 2015, 2:20 AM • 2 Y

Y by Adventure10, Mango247

Let $n\in \Bbb{N}, n \geq 4.$ Determine all sets $A = \{a_1, a_2, . . . , a_n\} \subset \Bbb{N}$ containing $2015$ and having the property that $|a_i - a_j|$ is prime, for all distinct $i, j\in \{1, 2, . . . , n\}.$

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#2 h May 29, 2015, 6:21 AM • 1 Y

Y by Adventure10

First, we cannot have three numbers in the set of the same parity. If we did, then their three pairwise differences would be even, so would all be $2$ , but clearly this is impossible. Thus, our set contains at most two odd and two even numbers, and in particular contains exactly four elements, two odd and two even. Furthermore, the two odd elements are $2$ apart, so exactly one of $2013$ and $2017$ are in the set along with $2015$ . For the moment let's assume it is $2013$ .

Our two even numbers are also $2$ apart, so are either both less than $2013$ or both greater than $2015$ . Let us assume the former. Then our set is $a,a+2,2013,2015$ for some appropriate even $a$ . Then $2011-a,2013-a,$ and $2015-a$ are all prime, but this can only happen if they are the primes $3,5,7$ . Thus, $a=2008$ , and we get the set $2008,2010,2013,2015$ . In a similar manner we can find exactly four sets, based on choosing either $2013$ or $2017$ , and on the evens being higher or lower than $2015$ .

They are $2008,2010,2013,2015$ ,

$2013,2015,2018,2020$ ,

$2010,2012,2015,2017$ ,

$2015,2017,2020,2022$ .

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#3 h May 29, 2015, 7:14 AM • 2 Y

Y by Adventure10, Mango247

good one

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#4 h Apr 8, 2025, 6:27 PM

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Let n \in \mathbb{N}, n \geq 4. Determine all sets
$A = \{a_1, a_2, \ldots, a_n\} \subset \mathbb{N}$ such that 2015 \in A and |a_i - a_j| is prime for all distinct i, j.

\textbf{Solution:}

Suppose n \geq 5. Since each number is either even or odd, by the Pigeonhole Principle, at least 3 elements must have the same parity. But the difference of two numbers with the same parity is even, and the only even prime is 2.

So all such differences must be 2. But it's impossible to have 3 distinct numbers where all pairwise differences equal 2. Hence, no 3 elements can share the same parity.

Therefore, at most 2 numbers can be odd and 2 can be even, so n \leq 4. But the problem says n \geq 4, so n = 4.

Now we find all such 4-element sets A with 2015 \in A, such that all differences are prime.

\textbf{Case 1: } a_1 = 2015

Try:
$A = \{2015, 2017, 2020, 2022\}$
$\begin{aligned} 2017 - 2015 &= 2 \\ 2020 - 2015 &= 5 \\ 2022 - 2015 &= 7 \\ 2020 - 2017 &= 3 \\ 2022 - 2017 &= 5 \\ 2022 - 2020 &= 2 \end{aligned}$
All differences are prime.

\textbf{Case 2: } a_2 = 2015

$A = \{2013, 2015, 2018, 2020\}$
$\begin{aligned} 2015 - 2013 &= 2 \\ 2018 - 2013 &= 5 \\ 2020 - 2013 &= 7 \\ 2018 - 2015 &= 3 \\ 2020 - 2015 &= 5 \\ 2020 - 2018 &= 2 \end{aligned}$
Valid set.

\textbf{Case 3: } a_3 = 2015

$A = \{2010, 2012, 2015, 2017\}$
$\begin{aligned} 2012 - 2010 &= 2 \\ 2015 - 2010 &= 5 \\ 2017 - 2010 &= 7 \\ 2015 - 2012 &= 3 \\ 2017 - 2012 &= 5 \\ 2017 - 2015 &= 2 \end{aligned}$
Valid set.

\textbf{Case 4: } a_4 = 2015

$A = \{2008, 2010, 2013, 2015\}$
$\begin{aligned} 2010 - 2008 &= 2 \\ 2013 - 2008 &= 5 \\ 2015 - 2008 &= 7 \\ 2013 - 2010 &= 3 \\ 2015 - 2010 &= 5 \\ 2015 - 2013 &= 2 \end{aligned}$
All differences are prime.

\textbf{Answer: } The four valid sets are:
$\begin{aligned} &\{2015, 2017, 2020, 2022\} \\ &\{2013, 2015, 2018, 2020\} \\ &\{2010, 2012, 2015, 2017\} \\ &\{2008, 2010, 2013, 2015\} \end{aligned}$

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#5 h Apr 8, 2025, 6:30 PM

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I am not able to Write in latex formet!
so it will be painful to read this solution.
this is a quite interesting problem.
I am really very confused when see this problem.
I am not really except how easy this problem!

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