mathmanman wrote: All positive integers are distributed among two disjoint sets $N_{1}$ and $N_{2}$ such that no difference of two numbers belonging to the same set is a prime greater than 100. Find all such distributions. Proposed by N. Sedrakyan If $n$ is in $N_{1}$ then $n+101$ must be in $N_{2}$ and $n+103$ must be in $N_{2}$. But then $n+2$ must be in $N_{1}$. If $m$ is in $N_{2}$ then $m+101$ must be in $N_{1}$ and $m+103$ must be in $N_{1}$. But then $m+2$ must be in $N_{2}$. So, if $1$ is in one of the to subsets (and it must be), then all the odd positive integers are in the same subset. But if $2$ is in the same subset too than all the numbers are in only one subset $\to$ contradiction. So, $2$ (and all the evens) are in the other subset. The answer is: We have only $2$ possibilities: 1) All the odds are in $N_{1}$ when all the evens are in $N_{2}$ 2) All the odds are in $N_{2}$ when all the evens are in $N_{1}$ Obviously, this distibution garantees that the difference between two numbers in the same subset will be even, so it cannot be prime greater than 100. I hope I was clear. Sorry for my bad English!