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Bmo2 1996 q1

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I remember this problem, as ghost walker has suggested, first split it into x,y>0 and the cases where x or y=0 (treat this bit as a special case) then show x and y are even in the more general case I think - its been a while since I've done it though.

Reply 21

Dog4444

Original post by Tobedotty

Do you remember, do you finish with Pell's or Pythagorean or what?

EDIT:
Oh, actually, difference of squares might be much better idea.

(edited 12 years ago)

Reply 22

gff

Original post by Dog4444

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

Spoiler

Can you convince me that these are the only solutions in those special cases?

The sum of an odd and an even integer is always an odd integer, and all odd integer squares are congruent to 1 modulo 4.

Reply 23

Dog4444

Right,
Working mod3
3^y is 0 mod 3
z^2 is 1 mod 3
For any y and z.
So 2^x is -1^x mod 3 therefore x is even.

Working mod 4, you get the same thing for y.

Is it solid?

Original post by gff

Can you convince me that these are the only solutions in those special cases?

Well, it's obvious? :s-smilie:

The only z which makes two brackets *number to the power of something, if n, would be larger, it's not possible for brackets to be two different powers of the same number?
Well, it just seems too obvious. Don't tell me, there are others, please. :biggrin:

(edited 12 years ago)

Reply 24

Blutooth

Original post by Dog4444

No, I haven't. You don't need to, if you can prove that (4,2) is the only non-trivial solution.

Also, I tried binomials.

2^x+(2+1)^y=2^x+2^y+a2^{y-1}+b2^{y-2}...+1=z^2

But I don't know what can I do further on.

Have a proof to the problem using that fact. Will type it up over the next hour if you haven't already solved it by then :smile:

(edited 12 years ago)

Reply 25

Dog4444

Original post by Blutooth

Have a proof to the problem using that fact. Will type it up over the next hour if you haven't already solved it by then :smile:

Thanks, may be solution in spoiler and hint first? :colondollar:

Reply 26

Blutooth

Original post by Dog4444

Thanks, may be solution in spoiler and hint first? :colondollar:

Actually, I got stuck at the point where

2^{x_0-1} - 1=3^{y_0}

. Need to find if there are solutions to that equation, and then we are done. I suspect there is only one :P. Essentially you just look at the binomial expansion mod 4 and then use a value of x or y you obtain there and check for solutions to z^2-2^2x.

Edit: looking at your first post, it seems I have got stuck at the same spot as you, using a different method to get there. Sorry I got your hopes up, will keep trying.

(edited 12 years ago)

Reply 27

gff

Original post by Dog4444

...

Some more details.

Spoiler

y = 0, x \geq 1

, then

2^x = (z - 1)(z + 1)

implies that

z \pm 1

are part of a geometric progression, but they're two consecutive terms.

Reply 28

Blutooth

Original post by gff

Some more details.

Spoiler

y = 0, x \geq 1

, then

2^x = (z - 1)(z + 1)

implies that

z \pm 1

are part of a geometric progression, but they're two consecutive terms.

Original post by Dog4444

Thanks, may be solution in spoiler and hint first? :colondollar:

OP, looking at your calculations and mine there seems to be one major difference. At the end I get a Pell equation of the form

(3^{y_1})^2- (2^{x_1-1})^2=1

This difference is because i factorised

z^2-3^x

rather than

z^2-2^y

, as you did. Pretty sure I have the soln. Will type it up soon.

(edited 12 years ago)

Reply 29

Dog4444

Original post by gff

Some more details.

Spoiler

y = 0, x \geq 1

, then

2^x = (z - 1)(z + 1)

implies that

z \pm 1

are part of a geometric progression, but they're two consecutive terms.

Does it mean my prove with mods isn't sufficient? And I can't follow the last sentence of first paragraph.

Original post by Blutooth

...

Thanks for trying, but it doesn't move anywhere. I'll try some algebraic manipulation.

Reply 30

Blutooth

I think this is what you are looking for MR OP :smile:

SOlution to the last part is here down below.

Spoiler

Only one further solution. We are done OP. Bullyacha.

(edited 12 years ago)

Reply 31

Dog4444

Original post by Blutooth

...

Sort of the same thing, what I do now.

[br]2^{2a}+3^{2b}=z^2[br]3^{2b}=(z-2^a)(z+2^a)[br]\gcd(z-2^a,z+2^a)=(z-2^a,2^{a+1})=1[br]z+2^a=3^{2b}[br]z-2^a=1[br]2^{a+1}=3^{2b}-1[br]2^{a+1}=(3^b-1)(3^b+1)[br]

Because brackets should be consecutive numbers of the form to the power of 2, and the difference between them is 2, the only valid b is 1. So y=2.

2^{a+1}=8[br]a=2, x =4, z=5[br]

How is it?

(edited 12 years ago)

Reply 32

Blutooth

Original post by Dog4444

Does it mean my prove with mods isn't sufficient? And I can't follow the last sentence of first paragraph.

Thanks for trying, but it doesn't move anywhere. I'll try some algebraic manipulation.

I think I've got what you are after spoilered above. If you come across some mistakes, I think I've fixed them. Will take a look at ur work now.

(edited 12 years ago)

Reply 33

Blutooth

Original post by Dog4444

Sort of the same thing, what I do now.

[br]2^{2a}+3^{2b}=z^2[br]3^{2b}=(z-2^a)(z+2^a)[br]\gcd(z-2^a,z+2^a)=(z-2^a,2^{a+1})=1 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![br]z+2^a=3^{2b}[br]z-2^a=1[br]2^{a+1}=3^{2b}-1[br]2^{a+1}=(3^b-1)(3^b+1)[br]

Because brackets should be consecutive numbers of the form to the power of 2, and the difference between them is 2, the only valid b is 1. So y=2.

2^{a+1}=8[br]a=2, x =4[br]

How is it?

The line with !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! is wrong isn't it? Generally when you come across a factorisation like that it's not the best way to deal with it. What if z=2^a+1 -2^a.

Do this z - 2^b=3^{a/m}
z + 2^b=3^{m}

(edited 12 years ago)

Reply 34

Dog4444

Original post by Blutooth

I think this is what you are looking for MR OP :smile:

Spoiler

2^{a/m+m}=2^a?

Reply 35

Dog4444

Original post by Blutooth

The line with !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! is wrong isn't it?

Don't think so at the moment.
Why do you think so?

Reply 36

Blutooth

Original post by Dog4444

2^{a/m+m}=2^a?

Sorry should be a minus instead of a division. bit tired, but if you just change that, and that alone in my workings the proof will still hold. Have corrected the error and the proof should hold now. Thanks for pointing it out.

(edited 12 years ago)

Reply 37

Blutooth

Original post by Dog4444

Don't think so at the moment.
Why do you think so?

What if

z=2^{a+1} -2^a.

(edited 12 years ago)

Reply 38

Dog4444

Original post by Blutooth

I think this is what you are looking for MR OP :smile:

SOlution to the last part is here down below.

Spoiler

But this is a contradiction, as both factors can't be 1.

Honestly, I can't follow it from "This has no solutions if a..." . If you're trying to prove that there is no solutions, when x,y>0. Try x=4,y=2,z=5.You get 16+9=25.

Original post by Blutooth

What if

z=2^{a+1} -2^a.

Do this

[br]z - 2^b=3^{a-m}[br]z + 2^b=3^{m}

Nearly made the mistake you pulled me up on again lol :P

I don't understand what you want to say here either.

Reply 39

Blutooth

Original post by Dog4444

Honestly, I can't follow it from "This has no solutions if a..." . If you're trying to prove that there is no solutions, when x,y>0. Try x=4,y=2,z=5.You get 16+9=25.

I don't understand what you want to say here either.

"gcd(z-2^a,z+2^a)=(z-2^a,2^{a+1})=1

This line of yours is wrong is what I was trying to say. Consider z=2^{a+1}-2^{1}.

I will write out a full solution tomorrow morning. And with regards to earlier I was aware of that solution actually. it occurs when a-m=1, but i'm pretty sure it is the only one of its kind- was just hurrying with the latex cos i got to sleep..