Prove or disprove that L = {$a^nb^m$ | $m ≠ 3n + 5$} is a regular

Question

How can I prove or disprove that $L = \lbrace a^nb^m$ | $m ≠ 3n + 5 \rbrace$ is a regular language?

Attempt
Assume $L$ is regular, then its complement $L^\complement$ is also regular.
$L^\complement = \lbrace a^nb^{3n+5} \mid n ∈ N \rbrace$.
By the pumping lemma, there exists a constant natural $p$. Let $s = a^pb^{3p+5}$ ∈ $L^\complement$. $|s| \geq p$ so there exists a split $s =xyz$ where $|x| \geq 1$, $|xy| \leq p$.
If I choose $i = 2$, then $xy^2z = a^{|x|}a^{2|y|}a^{p-|x|-|y|}b^{3p+5}=a^{p+|v|}b^{3p+5} \notin L^\complement$.
But I'm not sure how to continue to disprove that L isn't regular.

Answer 1

Assume $L$ is regular, then its complement $L^\complement$ is also regular.
Since $L^\complement$ is regular then $L^\complement$⋂ {$a^*b^*$} = $L_2$ is also regular.
$L_2 = \lbrace a^nb^{3n+5} \mid n ∈ N \rbrace$.
By the pumping lemma, there exists a constant natural $p$. Let $s = a^pb^{3p+5}$ ∈ $L_2$. $|s| \geq p$ so there exists a split $s =xyz$ where $|x| \geq 1$, $|xy| \leq p$.
If I choose $i = 2$, then $xy^2z = a^{|x|}a^{2|y|}a^{p-|x|-|y|}b^{3p+5}=a^{p+|v|}b^{3p+5} \notin L_2$.
This contradicts the pumping lemma, therefore $L_2$ isn't a regular language as well as $L$.