Proof: a³ - a is always divisible by 6 (2 of 2: Proof by exhaustion)

[freddyfozzyfilms]

the product of n consectutive integers will always be divisible by n factorial

[beccabattalio]

I watched this when I first woke up this morning, and I must say it was super satisfying. Love a nice, tidy little proof.

[wyboo]

a^3-a is the product of three consecutive integers:
multiples of 2 (even numbers) occur every other number, so either one or two of the products is even, so a^3-a is divisible by 2
multiples of three occur every third number, so exactly one of the three is divisible by 3 and thus a^3-a is divisible by 3
so a^3-a is divisible by both 2 and 3, and therefore also 2*3=6

[jacoblepley]

Wow... as a senior in high school finishing up calc AB... this proof thoroughly amazed me...

[pranavanil]

This question was eating me alive for weeks, thank you very much sir

[IM-mciw]

At 5:50, RHS is product of 3 consecutive numbers: 2k, 2k+1, 2k+2. One of them must be a multiple of 3. Therefore LHS must be both divisible by 2, as well as 3, therefore 6. QED.

[oyeajugando]

Sos un genio!! Me salvaste de caer depresión por Álgebra ❤

[KirilF]

I feel like I can hear him all day and not get tired XD
best teacher ever ;)

[cquirklesed]

Any three consecutive integers will contain a multiple of 2 and also a multiple of 3. Hence, the product P of the 3 consecutive integers must be a multiple of 2 and 3 => P is divisible by 2×3=6.

Same goes for any n consecutive integers, i.e. n consecutive integers will contain an integer that is a multiple of all of {1, 2, ...n} and hence their product must be a multiple of 1×2×...×n=n!

[mangeshpuranik]

Not just cube. I have just realised that this might hold true for any odd power, starting from 3 i.e. a^5, a^7 and so on!

Numbers are beautiful 🔥

[youssefwahba]

the pigeonhole principle can be applied to prove that any three consecutive integers will have a multiple of 3.
Since all elements of Z have a remainder of 0, 1, 2 when dividing by 3, and the three consecutive integers cannot have the same remainder, at least one of the three leaves a remainder of 0, thus at least one is a multiple of 3.

[Piglet]

a^3 - a = (a-1)a(a+1), I.e three consecutive integers

Among any three consecutive integers, there must be at least one even number (hence divisible by 2) and at least one number divisible by three.

Split into cases
1. WLOG if one of a-1, a or a+1 is divisible by 2 and 3 simultaneously, I.e. divisible by 6, then clearly the product (a-1)a(a+1) is divisible by 6 and we are done.

2. If none of a-1, a or a+1 is divisible by 6, then one must be a multiple of 3 and another must be a multiple of 2 . Therefore a factor of 3 and a factor 2 can be factored from the product (a-1)a(a+1), and hence overall divisible by 6

[skilz]

n = 1
1^3-1 = 0 which also equals -1 * 0 * 1 = 0 0/6 = 0
2^3-2 = 6 which also equals 1 * 2 * 3 = 6 6/6 = 1
3^3-3 = 24 which also equals 2*3*4 = 24 24/6 = 4
4^3-4 = 60 which also equals 3*4*5 = 60 60/6 = 10
5^3-5 = 120 which also equals 4*5*6 = 120 120/6 = 20
n, n+1, n+2, n+3, n+4 each raised to the third power and subtracted by n is a multiple six since 0, 1, 4, 10, and 20 are integers, therefore a^3-a is divisible by 6. We don't need any more examples due to the properties of modulo arithmetic. What more proof do you need?

[andyiswonderful]

Yay, I did this one in my head! No Alzheimer's yet.

You can use part induction, part exhaustion for less work:
((a+1)³-(a+1))-(a³-a)
=(a+1)³-a-1-a³+a
=a³+3a²+3a+1-a-1-a³+a
=3a²+3a
if a is even, a=2b: 3(2b)²+3(2b)=12b²+6b
if a is odd, a=2b+1:
Or you could just do the double induction, it's not too bad:
(3(a+1)²+3(a+1))-(3a²+3a)
=3a²+6a+3+3a+3-3a²-3a
=6a+6

[elis]

Wow, love this! You are great at explaining :D

[cyprienchabin]

Another way is by using Fermat's little theorem plus the Chinese Remainder Theorem.
With this method you can also prove that x^5-x is divisible by 30 for all x

[dineshmathsclasses]

Nice explanation, thank you for such a crystal clear proof.

[soba]

Me: You can change the odd expression with 2k -1 yeah?
My math teacher: *nO.*

[mathew_pang]

any pro here know what app he is using?