# A level maths proof Q

'Use algebra to prove that the square of any natural number is either a multiple of 3 or
one more than a multiple of 3'

So far I've got: All natural numbers are either odd or even. Odd: 2k + 1
Even: 2k

However I've tried squaring them and I'm not getting very far.
(edited 2 years ago)

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Original post by JaffaCakeBiscuit
'Use algebra to prove that the square of any natural number is either a multiple of 3 or
one more than a multiple of 3'

So far I've got: All natural numbers are either odd or even. Odd: 2k
Even: 2k + 1

However I've tried squaring them and I'm not getting very far.

You may need to revisit your definition of odd and even.
Original post by mqb2766
You may need to revisit your definition of odd and even.

sorry - I meant for them to be the other way around!! I did get this right in my workings on paper. But I am still struggling to get an answer.
Original post by JaffaCakeBiscuit
sorry - I meant for them to be the other way around!! I did get this right in my workings on paper. But I am still struggling to get an answer.

Couldn't resist :-).

Try representing a number in terms of (almost) a multiple of 3, then squaring.
Rather than the usual odd/even definition.
I'm struggling with this one as well - I've tried using 3n and 3n+1 instread of the normal odd/even definition, and I've reached

x^2 = 3(3n^2)
and
x^2=3(3n^2 +2n) + 1

I feel like I'm close, but I'm not sure where to go from here. Any advice?
(edited 2 years ago)
Original post by benjabrahamson
I'm struggling with this one as well - I've tried using 3n and 3n+1 instread of the normal odd/even definition, and I've reached

x^2 = 3(3n^2)
and
x^2=3(3n^2 +2) + 1

I feel like I'm close, but I'm not sure where to go from here. Any advice?

Sort of right, but what are you trying to prove? You want to show that the "end result" is a mutliple of 3 or one more than a multiple of 3.
The "end result" is the square of every natural number, so how are you representing every natural number?
Also, not sure what happens on the right hand side in bold.
(edited 2 years ago)
Original post by mqb2766
Sort of right, but what are you trying to prove? You want to show that the "end result" is a mutliple of 3 or one more than a multiple of 3.
The "end result" is the square of every natural number, so how are you representing every natural number?
Also, not sure what happens on the right hand side in bold.

Thanks for the response.

On the right hand side in bold, I started with x=3n+1, then squared to get x^2 = 9n^2 + 6n +1, which then factorises to get x^2 = 3(3n^2 + 2n) + 1. (I mistyped the 2n part as 2 before).

I'm still not really sure - do I have to somehow show that (3n^2 + 2n) and (3n^2) represent every natural number? I'm using x to represent that - something to do with that?
Original post by benjabrahamson
Thanks for the response.

On the right hand side in bold, I started with x=3n+1, then squared to get x^2 = 9n^2 + 6n +1, which then factorises to get x^2 = 3(3n^2 + 2n) + 1. (I mistyped the 2n part as 2 before).

I'm still not really sure - do I have to somehow show that (3n^2 + 2n) and (3n^2) represent every natural number? I'm using x to represent that - something to do with that?

You've started with 3n and 3n+1, is that all natural numbers covered?
A simple example will show its not quite there, but you're doing the right analysis.
hey guys. try using 6n and 6n+1
Original post by michael18756
hey guys. try using 6n and 6n+1

Why?
Original post by mqb2766
Why?

because then it works
Original post by michael18756
because then it works

Really? How does it work for all the numbers that aren't expressible in the form 6n and 6n+1?
Original post by davros
Really? How does it work for all the numbers that aren't expressible in the form 6n and 6n+1?

I was convinced by #11.
Original post by mqb2766
I was convinced by #11.

It would make a good exam answer for a collection of humorous anecdotes.

Actually it would probably be an A* in this year's teacher-assessed grades
Original post by davros
Really? How does it work for all the numbers that aren't expressible in the form 6n and 6n+1?

every number can be expressed in the form 6n and 6n+1 you silly goose
Original post by michael18756
every number can be expressed in the form 6n and 6n+1 you silly goose

Really? You mean numbers like 2, 3, 4, 5, 8 ,9, 10, 11, etc???
Original post by davros
Really? You mean numbers like 2, 3, 4, 5, 8 ,9, 10, 11, etc???

Don't be so dismissive - if he's right, I can see a (dis-)proof of the twin prime conjecture here...
Original post by DFranklin
Don't be so dismissive - if he's right, I can see a (dis-)proof of the twin prime conjecture here...

Good spot! Perhaps I need to revisit my juvenile attempts at the FLT in the light of this.
Original post by davros
Good spot! Perhaps I need to revisit my juvenile attempts at the FLT in the light of this.

TBF if you didn't realize that his n could be k/6 where k is a natural number, Im not surprised you're struggling with FLT.
(edited 2 years ago)
Original post by JaffaCakeBiscuit
'Use algebra to prove that the square of any natural number is either a multiple of 3 or
one more than a multiple of 3'

So far I've got: All natural numbers are either odd or even. Odd: 2k + 1
Even: 2k

However I've tried squaring them and I'm not getting very far.

consider cases for 3k, 3k+1 and 3k+2 because every number is one of them