Proving x is Even in 2^x+3^y=z^2 (1)

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Homework Statement


i want to prove that x is even in this equation

2^x+3^y=z^2 (1)

Homework Equations





The Attempt at a Solution



what i have so far is

(1) is congruent to mod3
2^x=z^2(mod3)
when z=1 and x=2 then
4=1(mod3)
so therefore x is even?

have i proved that x is even, or am i missing anything?
 
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You proved that if x=2 that it is even perhaps. There are two questions that should immediately come to mind. What can z squared be mod three be as z varies, and what can two to the x look like mod three as x varies? Calculating some small examples of each should suggest a general formula
 
so z^2 =1 (mod3) unless z=3m for any integer m
with 2^x
im not sure whether
2=1(mod3) or 2=2(mod3)
assuming 2=2(mod 3)
when x is odd
then 2^(2k+1)=2(mod3)
and when x is even
2^x=1(mod 3)

so I am not too sure where to go from here
 
So if x is odd, does 2^x and z^2 ever agree mod 3
 
if i was right with assuming 2^(2k+1)=2(mod3)
then no when x is odd it doesn't agree with z^2
 
oh so does that mean that x has to be even, is that how i prove it?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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