<s>[INST] Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!

Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$

Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^3+(k+1)^3+(k+2)^3+(k+3)^3 = 4b$ where $b$ is some integer. We need to show $4$ divides $(k+1)^3+(k+2)^3+(k+3)^3+(k+4)^3$.
\begin{align}
(k+1)^3 & +(k+2)^3+(k+3)^3+(k+4)^3\\ 
&= 4k^3+30k^2+90k+100\\
&=(k^3+(k+1)^3+(k+2)^3+(k+3)^3)+12k^2+48k+36\\
&=4b+12k^2+48k+36 \qquad \text{(by inductive hypothesis)}\\
&=4(b+3k^2+12k+9)
\end{align}
Since $b$ is an element of any integer this holds true for $(k+1)$. Hence proven.
 [/INST] To go from
$n$ to $n+1$,
you subtract $n^3$
and add $(n+4)^3$.
This means that
the sum changes by
$(n+4)^3 - n^3$,
so if this is divisible by 4,
divisibility by 4 remains.
But
$(n+4)^3 - n^3
=(n^3+12n^2+48n+64)-n^3
=12n^2+48n+64
=4(3n^2+12n+16)
$
is divisible by 4.
Since the first sum
(for n=0)
is
$0^3+1^3+2^3+3^3
=1+8+27
=36
$
is divisible by 4,
all are divisible by 4.
</s>