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Here is my proof:

Let $n \in \Bbb Z$. Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.

Without loss of generality (not sure if I can use this), let $n = 2k$.

Then, $n + 1 = 2k + 1$. $$\begin{align} (n + 1)^3 - n^3 & = (2k + 1)^3 - (2k)^3 \\ & = 8k^3 + 12k^2 + 6k + 1 - 8k^3 \\ & = 3(4k^2 + 2k) + 1 \end{align}$$

Let $m = (n + 1)^3 - n^3$.

Then, $m = 1 + 3(4k^2 + 2k) \Rightarrow m \equiv 1 \pmod 3$.

Therefore, $\forall m$, when divided by $3$, there remains a remainder $1$. So $(n + 1)^3 - n^3$ is not divisible by $3$.

Do I also have to prove this for the other case where $n$ is odd, or is this proof sufficient?

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It is not necessary to split into cases where $n$ is odd/even. Just use the binomial expansion of $(n+1)^3$, and note that the remainder of $3(n^2+n)+1$ upon division by $3$ will clearly be $1$ (the remainder would be $0$ if the difference were divisible by $3$). –  anon Feb 19 '13 at 21:25
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4 Answers

up vote 8 down vote accepted

$(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3(n^2 + n) + 1 \equiv 1$ (mod $3$)

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Thanks, seems like I've made my proof more complicated than it needs to be. –  icanc Feb 19 '13 at 21:30
    
I know the feeling. I've recently written a 3 page proof which I later realised can be done in two lines. –  muzzlator Feb 19 '13 at 21:36
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This can also be understood geometrically.

Arrange $n^3$ little cubes into a large cube of side length $n$, and then remove $(n-1)^3$ of them, leaving three flat sheets of cubes meeting at three edges.

Every cube left is either on the inside of one of the three sheets (and there are equally many of these on each sheet) or at one of the edges but not at the apex (again by symmetry each edge has equally many cubes in it), or it is the apex cube.

Therefore the total number of cubes left must be three times something, plus one cube at the apex.

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The difference from one cube to the next has a pattern the cubes 1^3 thru 10^3 are 1;8;27;64;125;216;343;512;729;1000 the differences are 7;19;37;61;91;127;169;217;271 they are all a multiples of 3+1 because they increase by multiples of 6 which would be multiples of three as well 7+12=19+18=37+24=61+30=91+36=127+42=169+48=217+54=271.... So if you start with a number not divisible by 3 and you add a number that is divisible by 3 then you will get another number not divisible by 3

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Let, number 1 = x then number 2 = x+1

(If Difference of cubes is divisible by 3 then if we mod that value by 3 it should give reminder as 0.)

Now, [Difference of cubes] mod 3 ,

= [(x+1)^3 - x^3] mod 3
= [x^3 + 3(x)(1)(x+1) + 1 - x^3] mod 3
= [3(x)(x+1) + 1] mod 3
= 1                         (because 3(x)(x+1) is always divisible by 3)

Thus, the reminder will be always 1.

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