Proof about an expression

[Thales12345]

A friend came up with the following question, but neither of us know the answer to it. Can someone help us?

GIVEN: a, b, c and d are natural numbers with a>b>c>d

ASKED: prove that (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) is always a twelvefold


We are thinking of this method for now: perhaps it is possible to prove that the factors in (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) are multiples of 2 and 3 (2*2*3 is the prime factorization of 12)?

 

[Otis]

A friend came up with the following question, but neither of us know the answer

Hi Thales. When you say "came up with", do you mean "made up"?

In other words, how do you know that the asked condition is actually true. ?

thinking of this … [some] factors in (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) are multiples of 2 and 3
2*2*3 is the prime factorization of 12

Did you experiment?

?

 

[Thales12345]

No, he didn't make up the question himself, but found it on the internet or in a book. So I assume that the asked condition is really true.


When we tried this, we found that for all natural numbers there are 4 options:
option 1) divisible by 2
option 2) divisible by 3
option 3) divisible by 2 and 3
option 4) not divisible by 2 or 3
Now these options should be tried at a, b, c and d (for example: what happens when option 1 - option 2?), but then there are 256 ways. So, there must be an easier way...

 

[Dr.Peterson]

A friend came up with the following question, but neither of us know the answer to it. Can someone help us?

GIVEN: a, b, c and d are natural numbers with a>b>c>d

ASKED: prove that (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) is always a twelvefold


We are thinking of this method for now: perhaps it is possible to prove that the factors in (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) are multiples of 2 and 3 (2*2*3 is the prime factorization of 12)?

I assume you take "a twelvefold" to mean "a multiple of 12". It is not a noun in my dialect, but it may be in yours.

One way to simplify the work is to define a-b = x, b-c = y, and d-c = z. Expression the product in terms of these three variables, and you'll have fewer cases.

You can also group cases. For example, start with what happens if x is even, and within that case consider whether y is even. You may, as you go, discover a way to collapse cases into fewer; or if you know modular arithmetic you may be able to state things more simply.

And after proving it is always a multiple of 4, you can move on to 3.

 

[Thales12345]

I assume you take "a twelvefold" to mean "a multiple of 12". It is not a noun in my dialect, but it may be in yours.

One way to simplify the work is to define a-b = x, b-c = y, and d-c = z. Expression the product in terms of these three variables, and you'll have fewer cases.

You can also group cases. For example, start with what happens if x is even, and within that case consider whether y is even. You may, as you go, discover a way to collapse cases into fewer; or if you know modular arithmetic you may be able to state things more simply.

And after proving it is always a multiple of 4, you can move on to 3.

When I express the expression in your way, I come to (x) (x + y) (x + y-z) (y-z) (-z).

When x is even, I don't know what 'happens' and how you determine whether y is even / odd.

If you want to prove that it is a multiple of 4, I just have to show that there are 2 even factors?
(A product of 2 even factors is always a multiple of 4)

How do you prove that it is a multiple of 3? I don't see it.

 

[Thales12345]

I just found out that there are 8 cases.
(o = odd, e = even)

1) x = e, y = e, z = e
->even factors: x, x+y, (x+y-z), y, y-z, -z
2) x = e, y = e, z = o
->even factors: x, x+y, y
3) x = e, y = o, z = e
->even factors: x, -z
4) x = e, y = o, z = o
->even factors: x, x+y-z, y-z
5) x = o, y = e, z = e
->even factors: y, -z, y-z
6) x = o, y = e, z = o
->even factors: y, x+y-z
7) x = o, y = o, z = e
->even factors: -z, x+y, x+y-z
8) x = o, y = o, z = o
->even factors: x+y, y-z

There are always at least 2 even factors, so the expression is always a multiple of 4.


Now we just have to prove that it is a multiple of 3. Can you help me?


Oh, and I know what's modular arithmetic, but I would not know how to use this here.

 

[Dr.Peterson]

When I express the expression in your way, I come to (x) (x + y) (x + y-z) (y-z) (-z).

When x is even, I don't know what 'happens' and how you determine whether y is even / odd.

If you want to prove that it is a multiple of 4, I just have to show that there are 2 even factors?
(A product of 2 even factors is always a multiple of 4)

How do you prove that it is a multiple of 3? I don't see it.

I meant to say z = c-d rather than d-c, which avoids the negatives; sorry. So (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) = (x)(x+y)(x+y+z)(y)(y+z)(z); you missed a factor here, but included y below.

I just found out that there are 8 cases.
(o = odd, e = even)

1) x = e, y = e, z = e
->even factors: x, x+y, (x+y-z), y, y-z, -z
2) x = e, y = e, z = o
->even factors: x, x+y, y
3) x = e, y = o, z = e
->even factors: x, -z
4) x = e, y = o, z = o
->even factors: x, x+y-z, y-z
5) x = o, y = e, z = e
->even factors: y, -z, y-z
6) x = o, y = e, z = o
->even factors: y, x+y-z
7) x = o, y = o, z = e
->even factors: -z, x+y, x+y-z
8) x = o, y = o, z = o
->even factors: x+y, y-z

There are always at least 2 even factors, so the expression is always a multiple of 4.

Good.

Now we just have to prove that it is a multiple of 3. Can you help me?


Oh, and I know what's modular arithmetic, but I would not know how to use this here.

You want to show that at least one of the six factors is a multiple of 3, that is, congruent to 0 (mod 3). Clearly this will be true if any of x, y, z = 0 (mod 3), so you again have 8 cases to consider (each of x, y, z being 1 or 2, mod 3).