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Suppose that $m$ and $n$ are positive integers such that every prime number which divides $n$ also divides $m$ (e.g. $n=25$ and $m=10$). How do you call such numbers? having the same "prime base" ? Thanks!

boaz
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    m= 10 and n= 25 are NOT an example of "every prime number that divides m also divides n". The prime number "2" divides 10 but does not divide 25. – user247327 Mar 11 '19 at 14:00
  • I thinks it was ok...any way, thanks for the comment @user247327. – boaz Mar 11 '19 at 14:02
  • does every prime number that divides m also have to divide n? If not then m is just a multiple of n. – cand Mar 11 '19 at 14:18
  • no, i mean that $m$ satysfies the following condition : $p\mid n\to p\mid m$ – boaz Mar 11 '19 at 14:26
  • It seems cand has the right terminology – Mikael Jensen Mar 11 '19 at 14:56
  • It's equivalent to: $n$ divides some power of $m$, or $,{\rm rad}(n)\mid m,,$ or $,{\rm rad}(n)\mid {\rm rad}(m)$ where $,{\rm rad}(n),$ denotes the radical of $,n = $ product of its distinct prime factors. Computing the radical is important in (algebraic) number theory, but there is no known fast algorithm. – Bill Dubuque Mar 11 '19 at 14:57
  • Your question is equivalent to "What do you call $m$ when $rad(n)\mid m$?" I think you just say $m\equiv 0 \mod{rad(n)}$. As I typed this, I see Bill Dubuque says about the same. – Keith Backman Mar 11 '19 at 15:00

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