If $n\geq 2$, then we must have $2^m-m \leq n^m-m \leq m^2+2m \implies 2^m \leq m^2+3m$, which means that $m$ can only be $1, 2, 3, 4, 5$. Testing cases yields the ordered pairs for $(m, n)$ as $(1, 2); (1, 4); (2, 2); (3, 2); (4, 2)$. If $n=1$, then we must have $m-1 | m^2+2m$, and solving this yields $m=2, 4$, yielding the additional pairs $(2, 1)$ and $(4, 1)$.

bzprules wrote: If $n\geq 2$, then we must have $2^m-m \leq n^m-m \leq m^2+2m \implies 2^m \leq m^2+3m$, which means that $m$ can only be $1, 2, 3, 4, 5$. Testing cases yields the ordered pairs for $(m, n)$ as $(1, 2); (1, 4); (2, 2); (3, 2); (4, 2)$. If $n=1$, then we must have $m-1 | m^2+2m$, and solving this yields $m=2, 4$, yielding the additional pairs $(2, 1)$ and $(4, 1)$. Why if $2^m \le m^2+3m$ then $m \in \{1;2;3;4;5 \}$ ?

Explain for me please !!

hello, plugging the numbers $m=1;2;3;4;5$ in the inequality $2^m\le m^3+3m$ you will see that this is true. Sonnhard.

shinichiman wrote: Why if $2^m \le m^2+3m$ then $m \in \{1;2;3;4;5 \}$ ? $ 64=2^6>6^2+18=54 $, and increasing speed of $ 2^m $ is faster than $ m^2+3m $, so if $ m \ge 6 $, then $ 2^m>m^2+3m $. Therefore, only $ m=1, 2, 3, 4, 5 $ satisfy this inequality in positive integers.

I hope this is correct. We have that $n > 1$ since $(m-1, m^2 + 2m)=1$. Now $n^m \ge 2^m \ge m$ so $n^m - m \ge 0$. So we have $m^2 + 3m - n^m \ge 0$; it's easy to check that this is untrue for $n \ge 5$ and is also easy to prove that for $n=4$ our only option is $m=1$ which holds the divisibility. For $n=3$ we have $m=1,2$ and none of them hold the divisibility. Finally for $n=2$ we have $m=1,2,3,4,5$ and testing them we have that $m=1,2,3,$ hold the divisibility; so in conclusion the pairs $(n,m)$ that hold the divisibility are $(4,1); (2,1), (2,2), (2,3)$ and we are done.

Find all $m,n$ postive integers such: $m^2+2m|n^{m}-m$