I'm a high school student, so I think my question will be an easy one. I would like to know if there is an easy way of approaching these Diophantine equations:
$x=\frac{1}{2^{a}-3}$
$x=\frac{2^{b}+3}{2^{a+b}-9}$
$x=\frac{2^{b+c}+3*2^{c}+9}{2^{a+b+c}-27}$
$x=\frac{2^{b+c+d}+3*2^{c+d}+9*2^{d}+27}{2^{a+b+c+d}-81}$
As you can see, they follow a pattern, and there are indefinitely many of them, but approaching these 4 would be enough. Another important thing is that x,a,b,c,d,...>=1.
My guess is that the only solution to these equations is x=1 and a,b,c,d,...=2. However, I'm having a hard time proving that.
If there is no approach, would it be possible to solve something by using lattices?
Thanks!