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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.

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  • $\begingroup$ 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. $\endgroup$
    – Daniel W. Farlow
    Commented Jul 17, 2015 at 0:38
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    $\begingroup$ 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. $\endgroup$
    – David
    Commented Jul 17, 2015 at 0:42
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    $\begingroup$ The induction step amounts to showing $(k+4)^3-k^3$ is divisible by $4$. This is obvious, since $a-b$ divides $a^3-b^3$. $\endgroup$
    – André Nicolas
    Commented Jul 17, 2015 at 0:45
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    $\begingroup$ 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. $\endgroup$
    – Cameron Williams
    Commented Jul 17, 2015 at 1:16
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    $\begingroup$ 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 $64$.) $\endgroup$
    – David K
    Commented Jul 17, 2015 at 1:20

2 Answers 2

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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 $K$.

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  • $\begingroup$ See this page for how to write math formulas on this site. $\endgroup$
    – Winther
    Commented Jul 17, 2015 at 1:06
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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.

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