If the solve the problem


If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1,(m, n \in N)$ then the

greatest common divisor of the least values of

$\mathrm{m}$ and $\mathrm{n}$ is___________


$\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m} / 2}=\left(\frac{1+\mathrm{i}}{\mathrm{i}-1}\right)^{\mathrm{n} / 3}=1$

$\Rightarrow\left(\frac{(1+i)^{2}}{2}\right)^{m / 2}=\left(\frac{(1+i)^{2}}{-2}\right)^{n / 3}=1$

$\Rightarrow(\mathrm{i})^{\mathrm{m} / 2}=(-\mathrm{i})^{\mathrm{n} / 3}=1$

$\Rightarrow \frac{\mathrm{m}}{2}=4 \mathrm{k}_{1}$ and $\frac{\mathrm{n}}{3}=4 \mathrm{k}_{2}$

$\Rightarrow \mathrm{m}=8 \mathrm{k}_{1}$ and $\mathrm{n}=12 \mathrm{k}_{2}$

Least value of $m=8$ and $n=12$

$\therefore \quad \mathrm{GCD}=4$

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