Proof that $a(a+1)(2a+1)$ is divisible by $6$ for every integer a

Question

This is from the book Elementary Number Theory by Jones & Jones

Example 3.6

Let us prove that a(a+1)(2a+1) is divisible by 6 for every integer a

By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2 or 3$. If $a \equiv 0$ then $a(a+1)(2a+1) \equiv 0 \cdot 1 \cdot 1 \equiv 0$, if $a \equiv 1$, then $a(a+1)(2a+1) \equiv 1 \cdot 2 \cdot 3 = 6 \equiv 0$, and similar calculations (which you should try for yourself) show that $a(a+1)(2a+1) \equiv 0$ in the other 4 cases, so $6 \vert a(a+1)(2a+1)$ for all a.

I don't understand the proof at all starting with the first line - By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2 or 3$. - How does taking absolute residues mod(6) give $a \equiv 0,\pm1,\pm2 or 3$?

Well these residues are the six consecutive integers with least positive sum (not the way it is usually put). If you are looking for the residues with least absolute value - one representative from each class - it is ambiguous whether you take $3$ or $-3$ as representative of that class. Generally this gets resolved in favour of the positive value unless there is a special reason otherwise. (I've not seen it used, but you could choose the value which makes the product of the non-zero representatives positive, for example). — Mark Bennet, Sep 23 '20 at 07:31
A quick (almost combinatorial) proof is the sum of the squares of the first $n$ positive integers. — Favst, Sep 23 '20 at 12:48

score 4 · Accepted Answer · edited Sep 23 '20 at 13:32

4

The remainder when you divide a number by $6$ can only be $0,1,2,3,4,5$.

Furthermore, $4 \equiv -2 \pmod{6}$ and $5 \equiv -1 \pmod{6}$.

After that, you just have to enumerate the $6$ cases to verify that it is true.

edited Sep 23 '20 at 13:32

J. W. Tanner

60,406

answered Sep 23 '20 at 07:14

Siong Thye Goh

149,520
20
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But how does taking absolute residues mod(6) give $a \equiv 0,\pm1,\pm2 or 3$? – user93353 Sep 23 '20 at 07:17
These are the possible residues with the smallest magnitudes. It has to satisfy $-\frac{6}{2} < r \le \frac{6}{2}$ – Siong Thye Goh Sep 23 '20 at 07:19
I am a beginner struggling with congruences for a month now. I really have no idea what you are on about. – user93353 Sep 23 '20 at 07:28
The set of remainders can be written as ${0,1,2,3,4,5}$. However, we can replace $4$ with $-2$ since $4-(-2) = 6$, similiarly we can replace $5$ by $-1$. – Siong Thye Goh Sep 23 '20 at 07:38
Got it - the part about the least absolute residues – user93353 Sep 23 '20 at 07:47

score 1 · Answer 2 · answered Sep 23 '20 at 07:37

Induction:

Check the validity for the base $a=1$. $$1(2)(3)$$ is divisible by 6. As it passes the first step, then we continues to the next step.
Assume the claim is valid for $a=k$. Or we can express as $$k(k+1)(2k+1)=6m$$
Check again for $a=k+1$. \begin{align} (k+1)(k+2)(2k+3) &= k(k+1)(2k+1) + 6(k+1)^2\\ &= 6m + 6 n\\ &= 6(m+n) \end{align}

which is obviously divisible by 6.

score 1 · Answer 3 · answered Sep 23 '20 at 08:17

I don't understand the proof at all starting with the first line - By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2 or 3$. - How does taking absolute residues mod(6) give $a \equiv 0,\pm1,\pm2 or 3$?

Let us prove that f(a) := a(a+1)(2a+1) is divisible by 6 for every integer a

By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2\ or\ 3$.

Shifting the standard system of residues (remainders) $\,0,1,\ldots 5\pmod{\!6}\,$ shows that any sequence $\,R\,$ of $\,6\,$ consecutive integers forms a complete system of residues (or remainders), i.e. every integer $\,a\,$ it is congruent to a unique $\,r_i\in R.\,$ Now $\!\bmod 6\!:\ a\equiv r_i\,\Rightarrow\, f(a)\equiv f(r_i)\,$ by the Polynomial Congruence Rule. Hence if we prove that $\,f(r_i)\equiv 0\,$ for all $\,r_i\in R\,$ then we can conclude that for all integers $a$ we have $\,f(a)\equiv 0,\,$ i.e $\,6\mid f(a).\,$ They prove $\,f(r_i)\equiv 0\,$ when $\,r_i = 1\,$ and leave to the reader the proofs for the remaining five elements of $R$.

score -1 · Answer 4 · answered Sep 23 '20 at 07:13

-1

Hint:

It's much better to use:

$$a(a+1)(2a+1)=a(a+1)(2(a-1)+3)=2\underbrace{(a-1)a(a+1)}_{\text{ The product of } 3\text{ consecutive integers }}+3a(a+1)$$

Use The product of $n$ consecutive integers is divisible by $n$ factorial

answered Sep 23 '20 at 07:13

lab bhattacharjee

274,582

But this doesn't answer my question - How does taking absolute residues mod(6) give $a \equiv 0,\pm1,\pm2 or 3$? – user93353 Sep 23 '20 at 07:17
@user93353, For any integer $2n,$ $$a\equiv 0,\pm1, \pm2, \cdots, n $$ – lab bhattacharjee Sep 23 '20 at 07:21
Where did 2n come into the picture from? – user93353 Sep 23 '20 at 07:23
@user93353, Here $$2n=6$$ – lab bhattacharjee Sep 23 '20 at 07:24
I have no idea what you mean. This is becoming like 20 questions – user93353 Sep 23 '20 at 07:26
+1 neutralized. It is a good hint for other visitors. No need to down vote. – Display Name Sep 23 '20 at 08:02

Proof that $a(a+1)(2a+1)$ is divisible by $6$ for every integer a

4 Answers4