So, I've seen on here this example:
$L_1 = \{a^n b^{2m} \mid n,m \geqslant 1 \}$
$L_2 = \{a^n b^{3n} \mid n \geqslant 0 \}$
$L_1 \cap L_2 = \{a^{2n} b^{6n} \mid n \geqslant 1 \}$
And I understood why: count(b) has to be divisible by 6, and count(a) is 3 times smaller than count(b).
But in this case:
$L_1 = \{a^n b^ma \mid n \geqslant 1, m \geqslant 0\}$
$L_2=\{a^n b^2 a^{2m} \mid n,m \geqslant 0 \}$
can we calculate the intersection as the last $a$ is odd in $L_1$ and even in $L_2$?