Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1)4^2+\dots+(4-1)4^{z-2}+(4-1)4^{z-1}$$
Which becomes:
$$4^z=1+4^1-4^0+4^2-4^1+4^3-4^2+\dots+4^{z-1}-4^{z-2}+4^z-4^{z-1}$$
After canceling everything out you are left with:
$$4^z=4^z$$
More generally:
$$a^z =1 + (a-1)\sum_{k=0}^{z-1} a^k$$
Therefore: $(2^x-1)/3$ will always be an integer when $x$ is even.
My question is: will there ever be an integer solution to $(2^x-1)/3$ when $x$ is odd?

If $x$ is odd, then $x=2k+1$ and then
$$2^{2k+1}-1=4^k\cdot 2-1\equiv1^k\cdot 2-1\equiv1\pmod 3$$
