Bmo2 1996 q1

http://www.bmoc.maths.org/home/bmo2-1996.pdf

Spoiler

But I don't how to prove that

a

has to be even (not sure if needed). And don't know what to do with the equation, it looks like the only solution is (a,b)=(2,1) (solutions, which can be represented as 3^2b and 2^2a), but I don't know how to prove it.
Help please. :biggrin:

Of course, if everything in the spoiler is correct. :facepalm2:

(edited 12 years ago)

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Reply 1

ghostwalker

You may wish to consider

2^3+3^0=3^2

in relation to your solution.

Reply 2

Dog4444

Original post by ghostwalker

You may wish to consider

2^3+3^0=3^2

in relation to your solution.

I'm not sure what do you want to say.
Yes, it's the simplest non-trivial answer to pell's equation.
Do you mean making recurrence relation? Or what?
I saw how to make recurrence on Wiki, but I was confused what to do to get actually x and y. Logs or something. Even so, I wouldn't know what to do further.

(edited 12 years ago)

Reply 3

ghostwalker

Original post by Dog4444

I'm not sure what do you want to say.

Perhaps I misunderstood your working. You seemed to be saying that x,y are even, and the example I posted contradicts that.

(edited 12 years ago)

Reply 4

Dog4444

Original post by ghostwalker

Perhaps I misunderstood your working. You seemed to be saying that x,y are even, and the example I posted contradicts that.

Oh, I see. Ignored the cases when x or y are zero. Thanks.

Reply 5

Dog4444

Original post by ghostwalker

Perhaps I misunderstood your working. You seemed to be saying that x,y are even, and the example I posted contradicts that.

Patched it.

Reply 6

ghostwalker

Original post by Dog4444

Patched it.

You might like to check that patching.

2^0+3^1=2^2

I felt there's something wrong with 2^0 :biggrin:

. Thanks, corrected.

Reply 8

ghostwalker

Original post by Dog4444

Corrected.

Once you get onto your first "Lemma", I don't see how you can come to that conclusion, that 2^x divides both those numbers (or that they are even integers). How do you justify it?

Reply 9

Dog4444

Original post by ghostwalker

Once you get onto your first "Lemma", I don't see how you can come to that conclusion, that 2^x divides both those numbers (or that they are even integers). How do you justify it?

Yeah, it's a typo. It should be 2.

(edited 12 years ago)

Reply 10

ghostwalker

Original post by Dog4444

Yeah, it's a typo. It should be 2.

Last one for today.

How do you know

\sqrt{3^y}

is an integer.

Reply 11

Dog4444

Original post by ghostwalker

Last one for today.

How do you know

\sqrt{3^y}

is an integer.

That's why y is even. It has to be an integer, otherwise it (z +/- sqrt3^y) won't be divisible by 2.

(edited 12 years ago)

Reply 12

ghostwalker

Original post by Dog4444

That's why y is even. It has to be an integer, otherwise it (z +/- sqrt3^y) won't be divisible by 2.

But you don't know that the factors are divisible by 2, you only know that the product is divisible by two.

(edited 12 years ago)

Reply 13

Dog4444

Original post by ghostwalker

But you don't know that the factors are divisible by 2, you only know that the product is divisible by two.

2 is a prime, right? None of these factors can be 1, so these two products should be in the form of 2^stuff, therefore they must be divisible by 2. It's very basic NT stuff.
Try to factorise any prime to some power, where none of these factors is 1. How any of them won't be divisible by this prime? Of course, if you work in positive integers. :confused:

(edited 12 years ago)

Reply 14

ghostwalker

Original post by Dog4444

My point is that you don't know those are integer factors.

2|5^2-\sqrt{3}^2

, but that doesn't imply

2|5\pm\sqrt{3}

(edited 12 years ago)

Reply 15

Dog4444

Original post by ghostwalker

My point is that you don't know those are integer factors.

2|5^2-\sqrt{3}^2

, but that doesn't imply

2|5\pm\sqrt{3}

Because 22=2*11? Do you have any examples when the number of the form prime to the power can be factorised, when none of these factors are one and not divisible by the prime?

EDIT: Yep, you're right mate. Never mind it's flawed inside out, I'll try other approach. Thanks.

(edited 12 years ago)

Reply 16

ghostwalker

Original post by Dog4444

Because 22=2*11? Do you have any examples when the number of the form prime to the power can be factorised, when none of these factors are one and not divisible by the prime?

If by factorised you mean split into integer factors, then no, none exist.

But I'm contesting the fact that you can split it into integer factors.
And no, I can't find a counterexample involving 2 and 3; but that doesn't prove they don't exist. Relaxing that restriction, we could use

2^5=7^2-\sqrt{17}^2

and

2\not|7\pm\sqrt{17}

Edit: Just saw you edit. Okey doke.

(edited 12 years ago)

Reply 17

Dog4444

Right,
The only approach I could think of is considering the equation as pythagorean triple (http://en.wikipedia.org/wiki/Pythagorean_triple):

a^2+b^2=c^2

Which has infinitely many solutions, but looks like (3,4) are the only ones, which can be represented in the form

(3^{0.5y},2^{0.5x})

But I don't know how to prove it.

Reply 18

ghostwalker

Original post by Dog4444

a^2+b^2=c^2

I presume you've found an alternative way of showing that x,y are even then (If not I can suggest something).

Yeah, don't know how you would progress beyond that.

Reply 19

Dog4444

Original post by ghostwalker

I presume you've found an alternative way of showing that x,y are even then (If not I can suggest something).

Yeah, don't know how you would progress beyond that.