In doing unrelated research, I conjectured that $n^3+(n+3m)^2$ is divisible by $3$ for all n in $[2+k,3+k]$ for all $k$ in $\mathbb{N}$ for all m in $\mathbb{N}$.
I'd like to prove this algebraically, but don't know how to frame the function for $n$.
I think that you wanted to say "for all $n$ in $[2 + 3k, 3 + 3k]$", since I can't see what purpose $k$ would have otherwise.
I will use congruence relations to prove your conjecture :
You have two cases :
$n \equiv 0 \mod 3$ (Meaning $n = 3 + 3k$):
Then $n^3 \equiv 0 \mod 3$.
Also, since $3m \equiv 0 \mod 3$, then $n + 3m \equiv 0 \mod 3$. Therefore, $(n + 3m)^2 \equiv 0 \mod 3$.
Summing our results, we get that $n^3 + (n + 3m)^2 \equiv 0 \mod 3$, which means that your conjecture is true for $n = 3 + 3k$.
$n \equiv 2 \mod 3$ (Meaning $n = 2 + 3k$):
Then $n^3 \equiv 8 \mod 3 \equiv 2 \mod 3$.
Also, since $3m \equiv 0 \mod 3$, then $n + 3m \equiv 2 \mod 3$. Therefore, $(n + 3m)^2 \equiv 4 \mod 3 \equiv 1 \mod 3$.
Summing up, we get that $n^3 + (n + 3m)^2 \equiv 2 + 1 \mod 3 \equiv 0 \mod 3$, which means that your conjecture is true for $n = 2 + 3k$.
In conclusion, your conjecture is true for all $n$ in $[2 + 3k,\,3 + 3k],\,k \in \mathbb{N}$.