<s>[INST] First proof writing practice: writing and style feedback, and is it correct? I'm not even sure if this is correct because there was no answer in the book and I'm new to proofs by induction. I'm mostly interested in feedback about the style and any proof-writing faux-pas'sss I may have made. It felt right to put the lemma up front to improve the flow of the main theorem and it possibly doesn't need to be proven, but I did so because I was advised you should write everything you need to to convince yourself in the early stages. Thanks

So I have to prove that $n(n^2 +5)$ is divisible by $6$ for all $n >=1$, but I start with a lemma I use later.
Lemma 1: $n^2 + n$ is even for all $n \in Z$.
Proof: Assume $n$ is even, then $n = 2m$ for some $m \in Z$, and:
$$n^2 + n = (2m)^2 + 2m = $$
$$ 4m^2 + 2m = $$
$$ 2(2m^2 + m) =$$
$$ n^2 + n = 2k$$
for some integer $k$. If $n$ is even, then $(n^2 + n)$ is also even.
Assume $n$ is odd and let $n = 2m -1$ for some $m \in Z$.
Thus,
$$n^2 + n = (2m -1)^2 + (2m -1)=$$
$$4m^2 - 4m +1 + 2m - 1 =$$
$$4m^2 - 2m  =2(2m^2 -1) = 2k$$ for some integer $k$ and again we find:
$$n^2 + n = 2k$$
Thus, $n^2 + n$ is even when $n$ is odd or even, proving $n^2 + n$ is even for all $n \in Z$.
Theorem: $n(n^2 + 5)$ is divisible by 6 for all integers $n>=1 $.
Proof:
By induction. Let $ A(n) = n(n^2 +5)$ and so:
$$A(1) = 1(1^2 + 5) = 6 = 6 \cdot 1$$
By assumption, $A(n) = 6k$ for some $k \in Z$:
$$ n(n^2 + 5) = (n^3 + 5n) = 6k $$
It suffices to show that $A(n+1) = 6r$ for some $r \in Z$:
$$A(n+1) = (n+1)[(n+1)^2 + 5]= $$
$$  (n+1)^3 + 5(n+1) =$$
$$  n^3 + 3n^2 + 8n + 6=$$
$$  n^3 + 3n^2 + (5n + 3n) + 6=$$
$$  (n^3 + 5n) + 3n^2 + 3n + 6=$$
By assumption,  $A(n) = (n^3 +5n) = 6k$ for some $k \in Z$ we have:
$$ 6k + 3n^2 +3n + 6 =$$
$$ 6k + 3(n^2 + n) + 6 =$$
By Lemma 1, we let $(n^2 +n) = 2j$ for some positive integer $j$:
$$ 6k + 3(2j) + 6 =$$
$$ 6k + 6j + 6 = $$
$$ 6(k + j + 1)$$
Let $ k + j + 1 = r$, then we have:
$$A(n+1) = 6r$$
Thus proving $A(1)$ is divisible by 6 and that if $A(n)$ is divisible by 6, then $A(n+1)$ is also divisible by 6.
 [/INST] Instead of editing I decided to make an answer, so you could compare better. In companion to the previous answers and comments, here is a shortened version of the results. Of course, a lot is personal taste how to write things down and you will learn a lot along the way.
Lemma 1. $n^2 + n$ is even for all $n \in \mathbb{Z}.$
Proof: As either $n$ or $n+1$ is even, it follows that $n^2+n = n(n+1)$ is even.
Theorem. $n(n^2 + 5)$ is divisible by 6 for all integers $n\geq 1 $.
Proof: Let $ A(n) = n(n^2 +5).$ We proceed by induction. Notice that $A(1)=6,$ this shows the induction base. Now assume that $A(n)$ is divisible by $6$ and consider
\begin{align*}
A(n+1) &= n^3 + 2n^2+6n+n^2+6 \\ &= (n^3+5n) + 3(n^2+n) +6\\ &= A(n) + 3(n^2+n) +6.
\end{align*}
The first and the second term are divisible by $6$ by the assumption and the lemma respectively, hence $A(n+1)$ is divisible by $6.$
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