
solve 2^x+3^y=z^2
basically i ahve got up to here my taking squares and modulus etc, if you need me to go into more detail i can. but i have basically got to
2^(k+1)=3^b1
where x=2k
this is meant to to me something, but im not sure what it is, any help would be much appreciated.
thanks

Re: solve 2^x+3^y=z^2
this is my solution so far
We are looking for solutions of:
2^x + 3^y = z^2 [1]
where x, y, and z are nonnegative integers.
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.
We now know x is even, let x=2k. So we have:
3^y=(z2^k )(z+2^k )
Now (z2^k ) and (z+2^k ) should both be powers of three, so therefore (z+2^k )(z2^k ) should equal a power of three, so
3^w=(z+2^k )(z2^k )
3^w≠2^2k=2^x
So 2^x should be divisible by three, but it isn’t. So let (z2^k )=3^a and (z+2^k )=3^b then we have
█((z+2^k )=3^b@(z2^k )=3^a )/(2^(k+1)=3^b3^a )
If a>0 then 3^b3^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^b1
Let a=k+1, then we have 2^a=3^b1.So this means that b≥0, otherwise 2^a wont be an integer, which means a≥1. Also a≤3 because if k>2, then 3^b+1≡0mod(8) which is a contradiction, so therefore a has to be either 1,2 or 3, making k=0,1 or 2.
When k=0 we have
2^1=3^b1 → 3=3^1 so therefore b=1 and
When k=1 we have
2^2≠3^b1 so therefore k=1 doesn’t exist.
When k=2 we have
2^3=3^b1 → 9=3^b so therefore b=2.
So the solutions of 2^x+3^y=z^2 are
X 0 4 3
Y 1 2 0
Z 2 5 3