well, one thing is to see that we don't lose any generality by assuming the numbers are distinct. why? becuase if two numbers are the same, say k is the repeated element, then k|k.

now, consider A - {2n+1,2n+2}. we can apply the induction hypothesis to that set. and if we have a divisible pair in that set, we have a divisible pair in A.

but....for this to be "legal", we must check an additional base case: n = 2.