# A question on divisibility 2

If, in my past question, $(2^{2m+1}-1)(2^{4m+2}+1)\mid (2^{2n+1}-1)(2^{4n+2}+1)$, then what is the relationship between $m$ and $n$?

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If $m=0$, $n$ arbitrary, the statement holds (left side is 5). –  coffeemath Dec 25 '12 at 21:06

If $2m+1\ |\ 2n+1$ then your divisibility statement will hold, since for $a$ and $b$ any odd natural numbers, $2^a-1 \ | 2^{ab}-1$ and also $4^a+1 \ | 4^{ab}+1.$ That is, if we write $$f(m)=(2^{2m+1}-1)(2^{4m+2}+1)=(2^{2m+1}-1)(4^{2m+1}+1)$$ and $2m+1\ |\ 2n+1$ , we will have $f(m)\ | f(n),$ for example $f(1)\ | f(4)$ follows from $2\cdot 1+1=3$ being a divisor of $2 \cdot 4+1=9$, and in fact $f(4)/f(1)=294409.$