No. $L$ isn't Regular. $L$ is union of two infinite languages, $a^{n+2}b^n, $ $a^nb^{n+2}.$
Proof using Myhill-Nerode Theorem:
To formally prove that $L$ is Non-regular, we can make show an infinite set of distinghuishable strings $S = \{ a^n | n \geq 1 \}.$ In given set $S,$ every two strings are distinguishable w.r.t $L,$ as proven below:
Take any two different strings in $L;$ $a^k,a^m$ where $m > k.$ For these strings, $b^{m+2}$ works as the distinghishing extension.
Since, an infinite set of distinghuishable strings exists w.r.t. $L$, so, $L$ is Not Regular (Myhill-Nerode theorem).
Proof using Pumping Lemma:
We will use Pumping lemma for regular languages to Prove that It is Not Regular.
Pumping lemma for Regular languages says
" If A languages is Regular,
then $\exists P \geq 1$, such that
$\forall \,\,strings\,\,w \in L$, Where $|w| \geq P$
$\exists x,y,z$, such that $w = xyz$
and $|xy| \leq P$
and $y \neq \in$ i.e. $|y| \geq 1$
and $\forall q \geq 0$, $xy^qz \in L$
In Words, If $L$ is regular, then there’s some magic number $P$(Called Pumping length). And if I take any string $w$ that is at least as long $P$, then I can break it up into three parts $x, y, $and $z.$ Now, the length of $xy$ is less than or equal to $P$, and also the length of $y$ is greater than or equal to $1$ and less than or equal to $P$.
NOTE that The first line is If $L$ is regular This means that our theorem only applies if the language is regular. It doesn’t apply if $L$ is not regular. I know you don't agree. But hold on. I'll prove it to you.
People say that We use the pumping theorem to show that a language isn’t regular. And It is indeed True. Then Why did I say the above line??
That's because Pumping lemma is a Proof by Contradiction. In other words, we assume $L$ is regular, then we show that it doesn’t satisfy the pumping theorem. This gives us a contradiction, so our initial assumption that $L$ is regular must be wrong.
So, Now let's consider the given language $L = \left \{ a^nb^k, |n-k| = 2 \right \}$
Let's Assume that $L$ is Regular.
Now, Since we have assumed that $L$ is Regular, so, This means that there’s some ”magic number” $P$ for the language $L$, such that the all the things in the rest of the theorem are true.
So, Let's take Pumping length as $P$ and now, Every String of length $\geq P$ must satisfy the rest of the conditions of the Pumping lemma.
So, as Pumping lemma states, For all $w$, where $|w| \geq P$, i.e. Sufficiently Large Strings, $w$ can be split into three Parts $x,y,z$.
So, Let's take some sufficiently large string with $|w| \geq P$ (as All the Strings with length greater than or equal to $P$ must satisfy the condition..)
So, Let $w = a^pb^{p-2}$ . Now let's try to split $w$ in three parts $x,y,z$ such that $|xy| \leq P$, $y \neq \in$
Now, $y$ has to be in the part $a^p$ because $|xy| \leq P $ and $|y| \geq 1$. You can try taking $y$ from $a$ to $a^P$ but You can find that $xy^qz$, $\exists q \geq 0$, Will Not be in the language.
So, We have shown that $L$ does not Satisfy the conditions of Pumping lemma. So, Our assumption that $L$ is Regular, was wrong. Hence, $L$ is Not regular.
$L$ is CFL. You can write it as Union of Two CFL languages as follows :
$\left \{a^nb^k:|n-k| = 2 \right \} = $ $\left \{a^nb^k:n-k = 2 \right \} $ $\cup \left \{a^nb^k:k-n = 2 \right \} $
You can find one more detailed explanation for another Non-regular language using Pumping lemma here in my answer : : https://gateoverflow.in/146637/prove-a-language-is-regular-using-pumping-lemma
