Proving 6 Divides a(a+1)(2a+1)

[Zhalfirin88]

Homework Statement

Prove that, for any integer a, 6 divides a(a+1)(2a+1)

The Attempt at a Solution

Well, 6 divides something if 2 and 3 divide the same number, so I must show that that product is even, and that the sum of its digits is divisible by 3. However, I don't see any algebraic manipulation to do either.

 

[Dickfore]

Hint: subtract the expression for $a \rightarrow a + 1$ and $a$. i.e. calculate:

$$(a + 1) (a + 2) \left[2 (a + 1) + 1\right] - a (a + 1) (2 a + 1) = ?$$

After you simplify this, what do you get?

Also, for $a = 1$, your expression is $1 \cdot 2 \cdot 3 = 6$, which is divisible by 6, and for $a = 0$, you get a zero, which is divisible by any non-zero integer.

It should be clear what method you should use by now.

Finally, what does your expression reduce to if you make the substitution $a \rightarrow -a$, i.e. what do you get for:
$$(-a) (-a + 1) \left[2 (-a) + 1\right]$$

and can you rewrite it in terms of your original expression for some positive integer value?

 

[Zhalfirin88]

Hint: subtract the expression for $a \rightarrow a + 1$ and $a$. i.e. calculate:

$$(a + 1) (a + 2) \left[2 (a + 1) + 1\right] - a (a + 1) (2 a + 1) = ?$$

After you simplify this, what do you get?

Also, for $a = 1$, your expression is $1 \cdot 2 \cdot 3 = 6$, which is divisible by 6, and for $a = 0$, you get a zero, which is divisible by any non-zero integer.

It should be clear what method you should use by now.

I have no idea what you're doing with that hint. I reduced it but I don't get why substituting $$(a+1)$$ and subtracting the expression for $$a$$ works?

Finally, what does your expression reduce to if you make the substitution $a \rightarrow -a$, i.e. what do you get for:
$$(-a) (-a + 1) \left[2 (-a) + 1\right]$$

and can you rewrite it in terms of your original expression for some positive integer value?

Okay, I reduced it to $$-2a^3+2a^2-a$$ but . . I'm not sure what you mean by rewriting it?

 

[Dickfore]

I reduced it but I don't get why substituting $$(a+1)$$ and subtracting the expression for $$a$$ works?

What did you get?

 

[Zhalfirin88]

$$6a^2+12a+6$$ Each term is divisible by 6, so it works, but I'm not really sure where you came up with subtracting expressions?

 

[Dickfore]

Let's call $f(a) \equiv a(a + 1)(2 a + 1)$. What you had shown is:

$$f(a + 1) - f(a) = 6 (a^{2} + 2 a - 1)$$

Now, if we assume that $f(a)$ is divisible by 6, what can you say about $f(a + 1)$? What principle does this remind you of?

 

[Zhalfirin88]

Kinda induction, but I'm guessing that's not what you're getting at?

If $$f(a+1)−f(a)=6(a^2+2a−1)$$ and we assume $f(a)$ is divisible by 6 then $f(a+1)$ is also divisible by 6.

 

[Dickfore]

Yes, I was precisely getting at that. But, what you had shown is that it is correct for positive integers (natural numbers). Now, you need to prove it for negative integers (for zero, I think I already showed it). That is why I told you to consider it for $a = -n$, where $n$ is again a positive integer. Do not expand it!

$$f(-n) = (-n) (-n + 1) \left[2(-n) + 1\right] = - n (n - 1) (2 n - 1) = (n - 1) \left[(n - 1) + 1\right] \left[2 (n - 1) + 1\right] = - f(n - 1)$$

That's it!

 

[Dick]

Not to distract from the excellent inductive approach, but you can also do this using modular arithmetic. The remainder when you divide a by 3 is either 0, 1 or 2. What does that make the remainder of a(a+1)(2a+1) in each of those cases?

 

[Zhalfirin88]

Thanks :)

 

[Zhalfirin88]

Not to distract from the excellent inductive approach, but you can also do this using modular arithmetic. The remainder when you divide a by 3 is either 0, 1 or 2. What does that make the remainder of a(a+1)(2a+1) in each of those cases?

If you did this by modular arithmetic you would have to take by mod 2 and mod 3, correct? And the remainder is zero =)

 

[Dick]

If you did this by modular arithmetic you would have to take by mod 2 and mod 3, correct? And the remainder is zero =)

Sure. But a(a+1) is obviously zero mod 2. mod 3 takes a little more work. Just a little.