Suppose $gcd(m,n)=d$ and $m=da,n=db$ where $gcd(a,b)=1$ Because $(a,b)=1$ so $(5b+a;ab)=1$ $5n+m|5m+n \iff 5b+a|5a+b\iff 5b+a|5ab+b^2=a(5b+a)-(a^2-b^2) \iff 5b+a|a^2-b^2$ $5b+a|25b^2-a^2 \implies 5b+a|24b^2$ $\iff 5b+a|24$ $\iff(a,b) \in \{(1;1);(3;1);(7;1);(19;1);(2;2);(12;2);(9;3);(4;4)\}$ $\implies b|a \iff n|m$

Same question 1 We have: $kb+a|k^2-1$ 2 is nice number If $k>2$ then we choose $a=2k-1$ and $b=k-2$ $\implies kb+a=k^2-1|k^2-1$ $\implies k-2|2k-1 \iff k-2|3 \iff k \in \{3;5\}$ Because 3 and 5 are nice numbers So 2;3;5 are all nice numbers