Question

Show that the product of any 3 consecutive positive integers is divisible by 6 using Fermat's infinite descent

Answer 1

What method is this :(

Answer 2

It seems to me like there is no useful way for infinite descent method in integer domain :(

The statement would look like this: "There is no smallest n for which the product is divisible." But it always drops down to standard divisibility proof - p|2 and p|3 -> p|6.

Answer 3

I do not know what method this is, but logically, if you have 3 consecutive integers, at least 1 of these numbers will be divisible by 2 and another will be divisible by 3, hence why they are also divisible by 6.

Answer 4

I mean if n is not divisible by 2 then n+1 is divisible by 2 and if n is also not divisible by 3 then n+2 is divisible by 3. If n is divisible by 3 then n+1 is divisible by 2.

Answer 5

@Frouzen

Yes I think any kind of proof with infinite descent must reach below step
"6 does not divide 1*2*3"
which is a contradiction, thus ending the proof.

Answer 6

@zimmah
It is easy to see that the product of any two consecutive integers is divisible by 2 because one of them is always even. But how do you prove divisibility by 3 ?

Answer 7

@rational Well, think about it, take any number (n) at all, and see if it divides by 3. If it doesn't then either n+1 or n+2 MUST divide by 3.

Answer 8

with number i mean integer.

Answer 9

odd and even numbers are usually described as 2k and 2k+1, k integers, where 2k+2 can be written as 2(k+1)
the same for division by 3:
3k
3k+1
3k+2
3k+3 = 3(k+1)
therefore the proof needs to be made for the three possibilities

Answer 10

Nice! :)

Frouzen is that for infinite descent ?

Answer 11

3k:
n(n+1)(n+2) = 3k(3k+1)(3k+2) = 3 [k(3k+1)(3k+2)
3k+1:
(3k+1)(3k+2)(3k+3) = (3k+1)(3k+2)·3(k+1) = 3 [(3k+1)(3k+2)(k+1)]
3k+2 simillary.

That is the standard proof of many divisibility problems. But I cannot see any logical way how to fit the infinite descent in...

Answer 12

Ahh yes division algorithm always works nicely for these divisibility problems when the modulus is small

Answer 13

I'm thinking something like this -
With the idea of arriving at a contradiction, let us assume that there exists a product of consecutive integers \(n(n+1)(n+2)\) that is not divisible by \(6\) :

\[6\nmid n(n+1)(n+2)\]
From this, we need to construct another smaller product of consecutive positive integers thats not divisible by \(6\). Since \(2\mid n(n+1)\) gives \(6\mid 3n(n+1)\) we have :
\[\begin{align}&6\nmid[n(n+1)(n+2)-3n(n+1)]\\
&6\nmid n(n+1)(n+2-3)\\
&6\nmid n(n+1)(n-1)\\
&6\nmid (n-1)n(n+1)\\
\end{align}\]
which is another smaller product of consecutive integers not divisible by \(6\).
repeating the whole argument we see that
\[\begin{align}&6\nmid (n-2)(n-1)n\\
&~~~~~~~\vdots\\
&~~~~~~~\vdots\\
&6\nmid 1\cdot 2\cdot 3\\
\end{align}\]
This is a contradiction because \(6\mid 6\), and ends the proof.

Answer 14

I would go for it this way. Actualy, I see no reason why not to go for (n-2)(n-1)n from the beginning as it is also a product of three consecutive integers, but this looks more... mathematically ... :D

Answer 15

I thought of using something else for 3n(n+1) but couldn't find anything more simpler... thanks for checking, appreciate it very much :)

Answer 16

So its sort of combination btw well ordering principles and proof by contradiction ??

Answer 17

I should learn that :)

Answer 18

Yes it is a contradiction proof using induction