I've found $ \boxed{n = 2^{1988}}$. Since $ m^n\equiv 1\pmod{2^{1989}}$ holds $ \ \forall m\in \mathbb{N}$, $ n$ has to be the $ \textrm{lcm}$ among every $ \textrm{ord}_{2^{1989}}(m_i)$ which means $ \big(n, \textrm{ord}_2^{1989}(m)\big)\not = 1$. Being $ \textrm{ord}_{2^{1989}}\vert \phi (2^{1989}) = 2^{1988}\Rightarrow n = 2^{a}, \ a\in \mathbb{N}$. Again to the congruence: $ m{}^2{}^a - 1\equiv \underbrace{(m - 1)(m + 1)(m^2 + 1)(m^4 + 1)\cdots (m{}^{2{}^{a - 1}} + 1)}_{2^a \ \textrm{factors}}\equiv 0\pmod{2^{1989}}$ If there exist two of this factors (except "$ m - 1$" which we cannot evaluate with the others) such that $ m{}^{2{}^{k}} + 1\equiv m{}^{2{}^{j}} + 1 \pmod{2^{1989}}$ or we get to the absurd $ m\equiv 0\pmod{2^{1989}}$ or for example, $ m{}^{2{}^i}\equiv 1 \pmod{2^{1989}}$. Now, since $ m{}^{2{}^i} - 1\vert m{}^{2{}^a} - 1$ we get what we wanted. We exclude now this case and suppose that all the remainder are different (always excluding "$ m - 1$"). We know that all the remainder will be even, since $ m$ is odd. So, all die possibile remainder are $ 2^{1988}$. To be sure that there is a "$ 0$" in those factors, $ 2^a\ge 2^{1988}$ so, $ n = 2^a = 2^{1988}$. Is it right?

let $ a$ be greatest number $ 2^a||m-1$. then prove that $ 2^{1889}||m^{1889-a}-1$.

you con refer to imo compendium book.if you can 't tell me ican write the solution for you.

orl wrote: Let $ m$ be a positive odd integer, $ m > 2.$ Find the smallest positive integer $ n$ such that $ 2^{1989}$ divides $ m^n - 1.$ I think that this problem statement is a bit ambiguous (or I am not understanding properly). I have also got the ans $ 2^{1988}$ (like EUCLA). But the solution in IMO compendium gives a bit general result. They says that $ 2^{1989 - s}$ if $ s \leq 1989$, and $ n = 1$ if $ s > 1989$ where $ s \in \mathbb{N}$ is the maximal integer satisfying $ m \equiv \pm 1\pmod { 2^s}$. Now my question is, which answer is correct? Though the solution of IMO compendium is more precise, I think that the problem just asks for the smallest $ n$ which works for every $ m$.

Moonmathpi496 wrote: I think that the problem just asks for the smallest which works for every $m$. Yes, this interpretation seems to be the correct one. But I don't think the answer is $2^{1988}$. Note that when $2^{1989}$ is replaced with $2^3$, then $n=2^1$ suffices. Similarly with $2^4$, then $n=2^2$ suffices. Then we can use induction: EDIT: Took me a while to realize that it wasn't actually induction If $m$ is odd, then $m^2=8x+1$ for some $x\in\mathbb{N}$. Thus \[m^{2^{1987}}=(8x+1)^{1986}\equiv 1\bmod{2^{1987}}\] from the Binomial Theorem. But I'm not sure how to show that this is the minimum such $n$...

Hint : Show that $3^{2^{n-3}} \equiv 1 + 2^{n-1} \pmod{2^n}$ by applying induction for $n \ge 3$.

You can use LTE when m=3. It yields the desired result.