# 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.

• If you did all of your algebra correctly, then I think it looks OK. It could be communicated more clearly (that is, it could be more polished), but the core elements are there in terms of correctness. Commented Jul 17, 2015 at 0:38
• Seems correct, but a suggestion to simplify: you don't need to expand the whole thing, because obviously$$(k+1)^3+(k+2)^3+(k+3)^3+(k+4)^3=\bigl( k^3+(k+1)^3+(k+2)^3+(k+3 )^3\bigr) +\bigl((k+4)^3-k^3\bigr)$$and you only need to expand the last bit. Commented Jul 17, 2015 at 0:42
• The induction step amounts to showing $(k+4)^3-k^3$ is divisible by $4$. This is obvious, since $a-b$ divides $a^3-b^3$. Commented Jul 17, 2015 at 0:45
• DO NOT delete stuff out of the body of your own post like this. It is bad form and against the spirit of the site. Commented Jul 17, 2015 at 1:16
• There is an algebra error in the proof. The term $36$ (which appears at the far right after two of the equality signs) should be $64$, so the $9$ at the end of the last equation should be $16$. The difference between the incorrect calculation and the correct calculation is itself a multiple of $4$, which is why the proof seems to work. (By the way, I spotted the error because there was reason to believe that $(k+4)^3 - k^3$ would appear somehow in the proof, and the constant term of that expression is $64$.) Commented Jul 17, 2015 at 1:20

Your answer seems right and looks like a short route, I would use modules to avoid the algebra but it wouldn't be induction. For $K \equiv 0 \pmod 4$ the modules of $4$ for the elements of the equation would be $0^3 + 1^3 + 2^3 + (-1)^3 \equiv 0 + 1 + 0 + (-1) = 0 \pmod 4$. And because there are always $4$ consecutive elements, the four module values are going to be the same for any $K$.
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.