nicetry007 wrote: $ m^{n}\equiv 1 (mod \; 2^{1989})$ and $ gcd(m,2^{1989}) = 1$. We have $ m^{\phi(2^{1989})}\equiv 1 (mod \; 2^{1989})$. Since $ \phi(2^{1989}) = 2^{1988}$, $ m^{2^{1988}}\equiv 1 (mod \; 2^{1989})$. Hence, $ n \leq 2^{1988}$. Let $ p$ be the smallest positive integer such that $ m^{p}\equiv 1 (mod \; 2^{1989})$. Claim: $ p \;| \;2^{1988}$. Suppose not. $ 2^{1988}= pq+r$ where $ 0 < r < p$. We have $ m^{2^{1988}}\equiv 1 (mod \; 2^{1989})$ $ m^{pq+r}\equiv 1 (mod \; 2^{1989})$ $ (m^{p})^{q}\cdot m^{r}\equiv 1 (mod \; 2^{1989})$ $ m^{r}\equiv 1 (mod \; 2^{1989})$. This contradicts the minimality of $ p$. Hence , $ p | 2^{1988}$. Let $ p = 2^{k}$. $ m^{p}-1 = m^{2^{k}}-1 = (m^{2^{k-1}}+1)\; (m^{2^{k-2}}+1)\; \cdots \;(m^{2}+1)\; (m+1)\;(m-1)$. Each of the $ m^{2^{l}}+1$ is divisible by 2 but not by 4. $ (m+1)\;(m-1) = m^{2}-1$ is divisible by 8 and there exist $ m$ for which $ m^{2}-1$ is not divisible by 16 (for example, $ m = 3$). Hence, the highest power of 2 that divides $ m^{p}-1 = m^{2^{k}}-1$ is $ 2^{k-1+3}= 2^{k+2}$. This implies $ k = 1987$. In other words, the smallest $ n$ for which $ 2^{1989}\; | \; m^{n}-1$ is $ 2^{1987}$.

Or we can use LTE: If $n$ odd then $v_2(m^n-1)=v_2(m-1) \ge 1989$, a contradiction. For example, $m=3$. Thus, $n$ is even. We get $v_2(m^n-1)=v_2(m-1)+v_2(m+1)+v_2(n)-1 \ge 1989$. Since for all odd positive integers number $m>1$ then $2^{1989}|m^n-1$ then we must have $v_2(n) \ge 1989$. It follows that $n=2^{1989} \cdot k$ with $k \in \mathbb{N}^*$. Thus, $n=2^{1989}$ is the smallest of $n$. More general: If $2^k|m^n-1$ then $2^k|n$ with $m$ is odd.

Min ord(m－1)＋ord(m＋1)－1 is $2$.So min n is $2^{1987}$.

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