For $k=1,2$: +) If $m \equiv 1 (\textrm{mod 4})$ then choose $n\equiv 1(\textrm{mod 4})$ +) If $m \equiv 3 (\textrm{mod 4})$ then choose $n\equiv 3(\textrm{mod 4})$ We define $n(k,m):n(k,m)^{n(k,m)}-m\vdots 2^k$ From above we solve for $k=1,2$ Now we will induction, for any $m$, assume exist $n(k,m)$. We will prove exist $n(k+1,m)$ +) If $n(k,m)^{n(k,m)}-m \vdots 2^{k+1}$ then choose $n(k+1,m)=n(k,m)$ +) If $n(k,m)^{n(k,m)}-m \equiv 2^k (\textrm{mod } 2^{k+1})$. We will choose $n(k+1,m)$ in the form $a:n(k+1,m)=a.2^k+n(k,m)$ We need $(a.2^k+n(k,m))^{a.2^k+n(k,m)}-m\vdots 2^{k+1}$, but $(a.2^k+n(k,m))^{a.2^k}-1 \vdots 2^{k+1}$ (LTE) So we need $(a.2^k+n(k,m))^{n(k,m)}-m\vdots 2^{k+1}$ Apply Newton"s binomial, we neew to choose $a: n(k,m).n(k,m)^{n(k,m)-1}.a2^k+n(k,m)^{n(k,m)}-m \vdots 2^{k+1}$ Which mean $2^k.(n(k,m).n(k,m)^{n(k,m)-1}.a+1) \vdots 2^{k+1}$. Just choose $a$ odd now Q.E.D