Bmo2 1996 q1

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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.

The line after that line is crucial, after that you are pretty much done.

Reply 41

Dog4444

Original post by Blutooth

"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..

I don't get where you got this z from.

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

Pre-last part by euclidean algorithm.
What part of it you think is wrong?

And yeah, lets' do it tomorrow (well, today). :biggrin:

Reply 42

Blutooth

Original post by Dog4444

I don't get where you got this z from.

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

Pre-last part by euclidean algorithm.
What part of it you think is wrong?

And yeah, lets' do it tomorrow (well, today). :biggrin:

That's not how you do the euclidean algorithm.

This is
if a>b
gcd(a,b)=gcd(a-b,b)

Also just considering that

z=2^{a+1}

, then

z-2^a=2^a

and

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

both of these have

2^a

or even 2 as a divisor and not 1. SO the GCD can't be 1.

It's a nice problem indeed. Yeah we'll get a better lid on it later 2day after some sleep :cool:

(edited 12 years ago)

Reply 43

Dog4444

Original post by Blutooth

That's not how you do the euclidean algorithm.

This is
if a>b
gcd(a,b)=gcd(a-b,b)

Also just considering that

z=2^{a+1}

, then

z-2^a=2^a

and

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

both of these have

2^a

or even 2 as a factor and not 1. SO the GCD can't be 1.

It's a nice problem indeed. Yeah we'll get a better lid on it later 2day after some sleep :cool:

\displaystyle \frac{z-2^a+2^{a+1}}{z-2^a}

remainder is

\displaystyle \frac{2^{a+1}}{z-2^a}

I think we need somebody else, who can look at everything from other perspective.

gcd(a,b)=gcd(a,b-a)
Yes, but it's just slower way to do this. Dividing is just much faster.

(edited 12 years ago)

Reply 44

Blutooth

Original post by Dog4444

\displaystyle \frac{z-2^a+2^{a+1}}{z-2^a}

remainder is

\displaystyle \frac{2^{a+1}}{z-2^a}

I think we need somebody else, who can look at everything from other perspective.

gcd(a,b)=gcd(a,b-a)
Yes, but it's just slower way to do this. Dividing is just much faster.

Here is your full solution OP. AFter the fact that x and y must be even.

we start off with (***)

2^{2a}+3^{2b}=z^2

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

For some integer g, we may write:

[br](1): z - 3^b=2^{2a}/g [br](2): z + 3^b=g [br]

But note from the first equation g must be of the form 2^m, otherwise the rhs a fraction, so we write

[br](1): z - 3^b=2^{2a-m} [br](2): z + 3^b=2^{m}

Note also that for a bit later:

m \geq 2a-m

[br](2)-(1)\Rightarrow 2^{m}-2^{2a-m}=2 \cdot 3^{b}

And Dividing by 2 yields.

(*): 2^{m-1}-2^{2a-m-1}= 3^{b}

The RHS is always odd . So the LHS must be Odd.
Note the LHS is odd as long as exactly one of

m-1

2a-m-1

is 0. As

m \geq 2a-m

, it must be the case that

2a-m-1=0

2a-m-1=0 \Rightarrow m=2a-1

Thus subbing for m in (*) , (*)becomes

3^{b}= 2^{2a-2}-1\Rightarrow 3^{2b}=(2^{2a-2}-1)^2

!!!!!!!!!!!

Going back to (***)

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

for some integer h we may write

[br](1): z - 2^a= 3^{2b}/h [br](2): z + 2^a=h [br]

But note h must be of the form h=2^K, otherwise we get the rhs of (1) being a fraction.

[br](1): z - 2^a=3^{2b-k} [br](2): z + 2^a=3^k [br]

Note therefore that:

k \geq 2b-k

[br](2)-(1)\Rightarrow 3^{k}-3^{2b-k}=2^{a+1}

And If the 2b-k>0 then the lhs is divisible by 3 but the rhs is not. Hence

2b-k=0

Subbing for k in

(2)-(1) \Rightarrow 3^{2b}-1=2^{a+1}\Rightarrow 3^{2b}=1+2^{a+1}

Combining this equation with the one marked !!!!!!!!!!!!! we get

3^{2b}=(2^{2a-2}-1)^2=1+2^{a+1}\Rightarrow[br]2^{4a-4}-2^{2a-1}+1=1+2^{a+1}\Rightarrow 2^{3a-4}-2^{a-1}=2[br]

Clearly a=2 is a solution. However, there are obviously no further solutions, quite simply from the fact that beyond a =2,

2^{3a-4}

grows faster than

2^{a-1}

And so the function

2^{3a-4}-2^{a-1}

is always increasing and gets much larger than 2.
At a=2, we have from (*)

3^{b}= 2^{2a-2}-1 \Rightarrow 3^{b}=3 \Rightarrow b=1

Finally we plug this into the first equation to verify that

2^{2a}+3^{2b}=z^2

. Clearly

2^{4}+3^{2}=5^2

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

(edited 12 years ago)

Reply 45

Dog4444

Original post by Blutooth

...

Seems fine to me. But I'm still not sure my gcd is wrong, at least I couldn't find a contradiction after putting some numbers in calculator.
Thanks, anyway.

Reply 46

Blutooth

Original post by Dog4444

\displaystyle \frac{z-2^a+2^{a+1}}{z-2^a}

remainder is

\displaystyle \frac{2^{a+1}}{z-2^a}

I think we need somebody else, who can look at everything from other perspective.

gcd(a,b)=gcd(a,b-a)
Yes, but it's just slower way to do this. Dividing is just much faster.

There is nothing wrong with this line. Your GCD method is fine, except for the last assertion that you make where you declare the gcd is 1. That is not the case

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

is 1 for all values of z, it may be 1 for some values of z. In fact the only values of z which will not make the gcd 1 are where z is an odd number. try a value of z that is a mutiple of 2.

Remember that the gcd is the greatest common factor of 2 numbers ie 2^(a+1)-2^(a) has a factor of 2^a.

(edited 12 years ago)

Reply 47

Dog4444

Original post by Blutooth

There is nothing wrong with this line. Your GCD method is fine, except for the last assertion that you make where you declare the gcd is 1. That is not the case