A set $S$ of positive integers is self-indulgent if $\gcd(a, b) = |a-b|$ for any two distinct $a,b\in S$
(a) Prove that any self-indulgent set is finite.
(b) Prove that for any positive integer n, there exists a self- indulgent set with at least n elements.
For the first question it can be seen that their will be no self indugent set consisting more no of elements twice of minimum element present. For sake of exploration i have picked any natural number. got its prime factorization. added every possible factors to the original number. it can be seen that for a self indulgent set where any arbitrary no N is minimum all the elements will be got by adding every factor of N initially. then there will be similar process to follow on the natural numbers we get by adding factors of N to N.Then for each of the numbers we will get a set of natural numbers.We will take intersection with all those sets with the first set we got. We can get multiple self indulgent set with minimum element N.
Now help me to proof that there will be a self indulgent set with n elements for any arbitrary natural number N. I can feel it is possible but just can't grasping it.
Here is the construction of self-indulgent sets of arbitrary finite cardinality given at quora.com (link in my comment on the question):
Let $a_1,a_2,\dotsc,a_n$ be a self-indulgent list of $n$ distinct positive integers.
Let $L$ be the least common multiple of $a_1,a_2,\dotsc,a_n$.
Then $L,L+a_1,L+a_2,\dotsc,L+a_n$ is a self-indulgent list of $n+1$ distinct positive integers.
For example; $1,2$ is a self-indulgent list, with $L=2$;
$2,3,4$ is a self-indulgent list, with $L=12$;
$12,14,15,16$ is a self-indulgent list, with $L=1680$;
$1680,1692,1694,1695,1696$ is a self-indulgent list of five distinct positive integers (and I'm too lazy to calculate $L$.)
The proof that this works is easy. We have $|(L+a_i)-(L+a_j)|=|a_i-a_j|=\gcd(a_i,a_j)$, and $\gcd(a_i,a_j)$ divides both $a_i$ and $a_j$, so it divides their lcm, which divides the lcm of $a_1,a_2,\dotsc,a_n$, which is $L$. Also, any $d$ which divides both $L+a_i$ and $L+a_j$ divides their absolute difference, which is $|a_i-a_j|$, which equals $\gcd(a_i,a_j)$, so $\gcd(L+a_i,L+a_j)=|a_i-a_j|$.