Problem Let $a,b$ be integers, and let $m,n$ be positive integers. Show that if
$a | (-b+2n)$ then $a \nmid (3b +7m)$ or $a \leq 6n+7m$.
Thoughts:
By contradiction, if $a | (3b+7m)$ and $a>6n+7m$.
If $a>(6n+7m)$ then $a \nmid 6n+7m$ since $m,n$ are positive.
I'm not really sure how to proceed. Hints appreciated.
Example:
Suppose $a=b=1$, and $m=n=2$, then $1$ does divide $3b+7m=3(1)+7(2)=17$ but $a=1 < 6(2)+7(2)=26$, ok this works.
The role of $m$ seems a bit unclear to me. For instance I could try to find a multiple of $(2b+7m)$ added to some multiple of $(6n+7m)$ that is divisible by $a$, that would somehow contradict that $a$ divides $(-b+2n)$, but not sure how.
Best Answer
Logically Equivalent to show the following: If $a|(-b +2n)$ and $a|(3b+7m)$, then $a \le (6n+7m)$.
So we show this next. To this end note the following: if $a|(-b+2n)$ then $a|3(-b+2n)=-3b+6n$. So $a$ divides both $-3b+6n$ and $3b+7m$, so $a$ divides the sum of these two integers which is $6n+7m$. But then if $a$ divides $6n+7m$ then $a$ must be no larger than $6n+7m$.