I'm looking for a way to prove that this system of equations has no solution for $A, B, C, a, b, c$ positive integers with $a, b, c$ distinct.
$$2^A * a = 3 * c + 1$$
$$2^B * b = 3 * a + 1$$
$$2^C * c = 3 * b + 1$$
If zero can be accepted, then there are solutions, for example: $A=0$, $B=3$, $C=2$, $a=13$, $b=5$, $c=4$.
I've checked solutions using a computer program, but I don't know how to start a mathematical proof.
Can someone help me?
Multiplying the equations together, $$abc 2^{A+B+C} = (3a+1)(3b+1)(3c+1)$$
so $$2^{A+B+C} = \left(3+\frac{1}{a}\right)\left(3+\frac{1}{b}\right)\left(3+\frac{1}{c}\right)$$
Since $a,b,c$ are distinct positive integers, the quantity on the right hand side is at most $$\left(3+\frac{1}{1}\right)\left(3+\frac{1}{2}\right)\left(3+\frac{1}{3}\right)=\frac{140}{3}$$
so $A+B+C \le 5$. Since $A,B,C$ are positive, there aren't many possibilities and it's easy to show these don't have integer solutions.