10

So this question has less to do about the proof itself and more to do about whether my chosen method of proof is evidence enough. It can actually be shown by the Principle of Mathematical Induction that the sum of the cubes of any three consecutive positive integers is divisible by 9, but this is not what I intend to show and not what the author is asking. I believe that the PMI is not the authors intended path for the reader, hence why they asked to prove divisibility by 3. So I did a proof without using the PMI. But is it enough?

It's from Beachy-Blair's Abstract Algebra Section 1.1 Problem 21. This is not for homework, I took Abstract Algebra as an undergraduate. I was just going through some problems that I have yet to solve from the textbook for pleasure.

Question: Prove that the sum of the cubes of any three consecutive positive integers is divisible by 3.

So here's my proof:

Let a $\in$ $\mathbb{Z}^+$

Define \begin{equation} S(x) = x^3 + (x+1)^3 + (x+2)^3 \end{equation}

So,

\begin{equation}S(a) = a^3 + (a+1)^3 + (a+2)^3\end{equation}

\begin{equation}S(a) = a^3 + (a^3 + 3a^2 + 3a + 1) + (a^3 +6a^2 + 12a +8) \end{equation}

\begin{equation}S(a) = 3a^3 + 9a^2 + 15a + 9 \end{equation}

\begin{equation}S(a) = 3(a^3 + 3a^2 + 5a + 3) \end{equation}

Hence, $3 \mid S(a)$.

QED

Bill Dubuque
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Derek W
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    It’s just fine, if you want to prove that $S(a)$ is divisible by $3$. – Brian M. Scott Apr 08 '13 at 00:09
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    Simpler: $\rm\ mod\ 3!:\ 0^3!+1^3!+2^3\equiv 0 + 1 + 8\equiv 0\ \ $ – Math Gems Apr 08 '13 at 00:10
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    Even simpler if you use $2\equiv -1$. – Berci Apr 08 '13 at 00:10
  • @Berci Right, but that might be slightly less clear to a beginner. – Math Gems Apr 08 '13 at 00:11
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    @MathGems: So does the modular arithmetic, if one hasn’t yet enountered it. – Brian M. Scott Apr 08 '13 at 00:12
  • The question that the author proposes is to prove that sum is divisible by 3. I have seen the question asking for the proof on the divisibility by 9 - which requires the PMI to show. I thought that the question was deliberately asking for 3 - hence this method of proof. – Derek W Apr 08 '13 at 00:13
  • @Brian Indeed. But being a problem from an abstract algebra textbook, there is a chance that the OP may know (or soon will know) modular arithmetic. – Math Gems Apr 08 '13 at 00:14
  • @MathGems: One would certainly hope so! (It actually comes in another $20$ pages or so, judging by the ToC.) – Brian M. Scott Apr 08 '13 at 00:20
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    Yes, I know modular arithmetic. I've taken courses in Discrete Math and Abstract Algebra. Plus I minored in CS as an undergrad. My first instinct is never the modular arithmetic approach - although it yields a very nice, quick, concise proof. – Derek W Apr 08 '13 at 00:22
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    @Derek For divisibility problems your first instinct should be modular arithmetic. It often simplifies such problems, because working with equations (congruences) is usually simpler than working with relations (divisibility). – Math Gems Apr 08 '13 at 00:51
  • The change of variable from $x$ to $a$ buys you nothing. – Kaz Apr 08 '13 at 06:48
  • @Kaz From my post: "But is it enough?" I'm asking if my proof - one foregoing the PMI approach is sufficient enough evidence. – Derek W Apr 08 '13 at 07:40

8 Answers8

8

Your solution is fine, provided you intended to prove that the sum is divisible by $3$.

If you intended to prove divisibility by $9$, then you've got more work to do!

If you're familiar with working $\pmod 3$, note @Math Gems comment/answer/alternative. (Though to be honest, I would have proceeded as did you, totally overlooking the value of Math Gems approach.)

amWhy
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4

Your solution is perfect.

If you are familiar with modular arithmetic, there are even much faster proofs, see the comments.

Or, instead of $x,x+1,x+2$, you could start out from $x-1,x,x+1$. But no more evidence is needed than yours.

Berci
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2

You have actually done enough work to show that the sum of the $3$ cubes is divisible by $9,$ not juat by $3,$ but you haven't explained that step: Note that (mod $9$), $S(a)$ is the same as $3(a^{3}+5a).$ But $a^{3}-a$ is divisible by $3$ for all integers $a,$ as you can see by checking $(3b-1)^{3}$ and $(3b+1)^{3}$ for integers $b.$ Hence $S(a) = 18a +9b$ for some integer $b.$

Geoff Robinson
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2

As was already mentioned, modular arithmetic is the most efficient way to solve this problem, but, if you really want to avoid it, you can still get by with slightly easier computation (smaller coefficients). Introduce a name, say $y$, for the middle one of the three integers, rather than the smallest. So you'd add $(y-1)^3+y^3+(y+1)^3$, which leads to a fair amount of cancellation and no coefficients bigger than 6.

Andreas Blass
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1

Any 3 consecutive numbers will always be of the form 3k-1,3k,3k+1.So 3k-1 ≡ -1mod 3. Where mod a ≡ x mod y means that x is the reminder obtained when a is divided by y.(3k-1)^3 ≡ -1 mod 3 . Similarly, 3k ^3 ≡ 0 mod 3. (3k+1)^3 ≡ 1 mod 3. Therefore (3k-1)^3 + (3k)^3 + (3k+1)^3 ≡ -1 + 0 + 1 mod 3 ≡ 0 mod 3. Hence it is always divisible by 3.

Manoj
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1

I like the three consecutive integers to be $k -1$, $k$,$\, k + 1 $, the sum of whose cubes are $3k(k^2 + 2)$, but that is just a matter of taste.

Hakim
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Ed Gray
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0

The set (unordered) of three consecutive numbers in module 3 is { 0, 1, -1 }

(-1)^3 = -1 (mod 3) [ equivalent: 2^3 = 2 (mod 3) ]

(0)^3 = 0 (mod 3)

(1)^3 = 1 (mod 3)

The sum of the three cubes is 0 (mod 3).

Angel Moreno
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0

Your approach and deduction is absolutely fine Just for the sake of completeness, here is an alternative prove

Known

$$(a+b+c)^3 = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 a^{2} c + 3 a b^{2} + 3 a c^{2} + 3 b^{2} c + 3 b c^{2} + 6 a b c $$ $$ = a^{3} + b^{3} + c^{3} + 3f(a,b,c)$$

Let $a=k-1,b=k,c=k+1$,then, $a,b,c$ represents consecutive numbers

So we have $$(a+b+c)^3=(k-1+k+k+1)^3=(3k)^3=27k^3$$ Thus $$a^{3} + b^{3} + c^{3} = (a+b+c)^3 - 3\cdot f(a,b,c) = 27k^3-3\cdot f(a,b,c)$$ $$\Rightarrow a^{3} + b^{3} + c^{3}=3\cdot(9k^3-f(a,b,c))$$

Abhijit
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