If $${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$$, (m, n $\in$ N) then the greatest common divisor of the least values of m and n is _______ .

$${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$<br><br>$\Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$ = i⁴ [ As i⁴ = 1]<br><br>$\Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$<br><br>$\Rightarrow$ m = 8 k₁ and n = 12 k₂<br><br>Least value of m = 8 and n = 12.<br><br>$\therefore$ GCD (8, 12) = 4