The line after that line is crucial, after that you are pretty much done.(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.
Bmo2 1996 q1


(Original post by Blutooth)
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 am=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..
Prelast part by euclidean algorithm.
What part of it you think is wrong?
And yeah, lets' do it tomorrow (well, today). 
(Original post by Dog4444)
I don't get where you got this z from.
Prelast part by euclidean algorithm.
What part of it you think is wrong?
And yeah, lets' do it tomorrow (well, today).
This is
if a>b
gcd(a,b)=gcd(ab,b)
Also just considering that , then and
both of these have 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 
(Original post by Blutooth)
That's not how you do the euclidean algorithm.
This is
if a>b
gcd(a,b)=gcd(ab,b)
Also just considering that , then and
both of these have 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
I think we need somebody else, who can look at everything from other perspective.
gcd(a,b)=gcd(a,ba)
Yes, but it's just slower way to do this. Dividing is just much faster. 
(Original post by Dog4444)
remainder is
I think we need somebody else, who can look at everything from other perspective.
gcd(a,b)=gcd(a,ba)
Yes, but it's just slower way to do this. Dividing is just much faster.
we start off with (***)
For some integer g, we may write:
But note from the first equation g must be of the form 2^m, otherwise the rhs a fraction, so we write
Note also that for a bit later:
And Dividing by 2 yields.
The RHS is always odd . So the LHS must be Odd.
Note the LHS is odd as long as exactly one of or is 0. As , it must be the case that
Thus subbing for m in (*) , (*)becomes
!!!!!!!!!!!
Going back to (***)
for some integer h we may write
But note h must be of the form h=2^K, otherwise we get the rhs of (1) being a fraction.
Note therefore that:
And If the 2bk>0 then the lhs is divisible by 3 but the rhs is not. Hence
Subbing for k in
Combining this equation with the one marked !!!!!!!!!!!!! we get
Clearly a=2 is a solution. However, there are obviously no further solutions, quite simply from the fact that beyond a =2, grows faster than And so the function is always increasing and gets much larger than 2.
At a=2, we have from (*) Finally we plug this into the first equation to verify that . Clearly
Only one further solution. We are done OP. Bullyacha. 
(Original post by Blutooth)
...
Thanks, anyway. 
(Original post by Dog4444)
remainder is
I think we need somebody else, who can look at everything from other perspective.
gcd(a,b)=gcd(a,ba)
Yes, but it's just slower way to do this. Dividing is just much faster.
Remember that the gcd is the greatest common factor of 2 numbers ie 2^(a+1)2^(a) has a factor of 2^a. 
(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 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. 
(Original post by Dog4444)
I suggest you wanted to write gcd will be 1 when z is odd? Well, z is always odd, isn't it? That's why it's 1.
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