Question

im meant to solve 2^x+3^y=z^2 and have got to 2^a=3^b-1, what does this tell me?

Answer 1

oh and a=k+1 where x=2k

Answer 2

Do you know anything about b?

Answer 3

We now know x is even, let x=2k. So we have:
3^y=(z-2^k )(z+2^k )
Now (z-2^k ) and (z+2^k ) should both be powers of three, so therefore (z+2^k )-(z-2^k ) should equal a power of three, so
3^w=(z+2^k )-(z-2^k )
3^w≠2^2k=2^x
So 2^x should be divisible by three, but it isn’t. So let (z-2^k )=3^a and (z+2^k )=3^b then we have
█((z+2^k )=3^b@(z-2^k )=3^a )/(2^(k+1)=3^b-3^a )
If a>0 then 3^b-3^a will be a power of three and 2^(k+1) is not divisible by three, so let a=0. We are left with
2^(k+1)=3^b-1

Answer 4

Hold on. Why do we know x is even?

Answer 5

So z^2 ≡1 (mod 3) unless z^2=3m for any integer m, and if x is odd, say x=(2k+1) for some integer k. However when z=3, we have 2^3+3^0=3^2, so one solution is when x=3,y=0 and z=3 When x is odd 2^(2k+1)≡2(mod 3) and when x is even 2^x≡1 (mod3) so z^2 is congrgent to 2^x when x is even.

Answer 6

assuming that is correct

Answer 7

What do you mean by "solve"?

Answer 8

find solutions for x y and z where thsi equality holds

Answer 9

In integers....

Answer 10

yes sorry

Answer 11

Number theory/Diophantine equations, right?

Answer 12

yes i think so, apparently 2^a=3^b-1 is meant to tell me something or maybe represent something, but im not sure what

Answer 13

Well it gives you a relation between a and b and therefore to k.

This is a subfield of Diophantine Equations called Exponential Diophantine Equations and there is no general theory covering them, you just play it by ear, follow your nose, whatever.

Answer 14

ok, never heard of that. any ideas on how i shoudl go about finishing it then?

Answer 15

I didn't actually read all the arguments above but if there is aquery about odd/even, probably best to it by cases.

Answer 16

It looks like you're trying to evaluate mod 3 only. It might also be helpful to valuate mod 2.

Answer 17

what should i be evaluating at mod 2?

Answer 18

Well, both sides of the equation. You can tell that \(z^2\) is even if and only if \(x=0\). So fix \(x=0\) and see if that gets you anywhere.

Answer 19

i think i did that at the beginning, because z is odd unless x=0, but thas only one solution

the solutions i have so far are

X 0 4 3
Y 1 2 0
Z 2 5 3

however im not sure if there are more after these

Answer 20

I tend to think those are the only ones, but I'm not sure how you would do that.

Answer 21

yer that's the trouble im having haha