Let $\frac{5n+m}{5m+n}=k$ $(5-k)n=(5k-1)m$ As RHS is positive LHS should be positive. So, $5-k>0$ $k=1,2,3,4$ $k=1\implies m=n\implies m|n$ $k=2\implies n=3m\implies m|n$ $k=3\implies n=7m\implies m|n$ $k=4\implies n=19m\implies m|n$

@above nice. Alternate solution. $5m+n|5n+m\implies 5m+n|24m$. Now assume to the contrary that $m\nmid n$. Let $n=mk+r$ where $0<r<m$. So, $$5m+n|24m\implies m(5+k)+r|24m\implies m|r$$which is not possible as $0<r<m$, hence a contradiction arises. Hence, $n|m$.

amar_04 wrote: @above nice. Alternate solution. $5m+n|5n+m\implies 5m+n|24m$. Now assume to the contrary that $m\nmid n$. Let $n=mk+r$ where $0<r<m$. So, $$5m+n|24m\implies m(5+k)+r|24m\implies m|r$$which is not possible as $0<r<m$, hence a contradiction arises. Hence, $n|m$. @above why this ...if $24 = c [m(5+k) +r ]$ ...then?

5m + n | 5n + m so there exists a k such that: (5m + n)k = 5n + m m(5k-1) = n(5-k) ---> k < 5 case 1 : k = 1 ---> 4m = 4n ---> m=n ---> m|n case 2 : k = 2 ---> 9m = 3n ---> 3m = n ---> m|n case 3 : k = 3 ---> 14m = 2n ---> 7m = n ---> m|n case 4 : k = 4 ---> 19m = n ---> m|n