# A level maths proof Q

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'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.

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.

Last edited by JaffaCakeBiscuit; 1 month ago

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#2

(Original post by

'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.

**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.

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(Original post by

You may need to revisit your definition of odd and even.

**mqb2766**)You may need to revisit your definition of odd and even.

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#4

(Original post by

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.

**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.

Try representing a number in terms of (almost) a multiple of 3, then squaring.

Rather than the usual odd/even definition.

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#5

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?

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?

Last edited by benjabrahamson; 1 month ago

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#6

(Original post by

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=

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

**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?

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.

Last edited by mqb2766; 1 month ago

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#7

(Original post by

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

Also, not sure what happens on the right hand side in bold.

**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.

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?

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#8

(Original post by

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 +

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?

**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?

A simple example will show its not quite there, but you're doing the right analysis.

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#12

(Original post by

because then it works

**michael18756**)because then it works

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#13

(Original post by

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

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

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#14

(Original post by

I was convinced by #11.

**mqb2766**)I was convinced by #11.

Actually it would probably be an A* in this year's teacher-assessed grades

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#15

**davros**)

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

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#16

(Original post by

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

**michael18756**)every number can be expressed in the form 6n and 6n+1 you silly goose

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#17

(Original post by

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

**davros**)Really? You mean numbers like 2, 3, 4, 5, 8 ,9, 10, 11, etc???

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#18

(Original post by

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

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

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#19

(Original post by

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

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

Last edited by mqb2766; 1 month ago

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#20

(Original post by

'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.

**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.

1

reply

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