I can construct DFA for $a^{3n+1}$ but pumping lemma says it is not regular Recently I stumbled across this language $L=\{a^{n}a^{{(n + 1)}^2-n^2} \in \Sigma^* \mid n\geq 0\}$ that I can rewrite as $a^{3n + 1}$.
So I applied the pumping lemma to see if it is non regular, and I get:

given a pumping length p I should have a string $z\in L$ such that $\mid z \mid \ge p$
I choose $z=a^pa^pa^pa$ such that $z\in L$ and $\mid z \mid = (3p + 1)\ge p$
now I can slipt the string in 3 parts, $z=uvw$ with $u=a^{p-1}, v=a,w=a^{2p}a$ and it satisfies the two following conditions of the lemma $\mid uv \mid \leq p$ and $\mid v \mid &gt; 0$
then given $z = uv^iw$, if I pump $v$ I see that $z \notin L$ because if for example I choose $i=2$ than $\mid z \mid = (p-1)+i+2p+1=3p+2 \geq 3p+1$

So $z \notin L$ and it means that $L$ is not regular, but I noticed that I can construct a DFA that accepts this language:

So now I'm really confused. Probably I'm applying the pumping lemma in the wrong way but I don't understand where.
UPDATE
Now, thanks to all of you, I think I understood my mistake. So to prove that a language is not regular I have to show that every decomposition fails to satisfy the pumping condition for some pumping number $i$. But in this case there is a decomposition that satisfies the pumping condition so I can't prove that the language is regular (but we know it is regular because we can construct a DFA that accepts it).
You should notice that if the conditions hold, the language can be regular (necessary, but not sufficient). However, you proved the third condition is not held in your case, hence it is not regular which this is not true. Therefore, your proof is not correct. 

On the other hand, if we suppose $x = a$ and $y = a^{3}$ and $z = 3n-4$ we can say $xy^nz$ is in $L$ for all $n\geq1$ ($p = 4$). Hence, we can't proof $L$ is not regular by pumping lemma.