On an Exponential Diophantine Equation of the form $g^y = ax + b$

Question

Consider the Exponential Diophantine equation:

$$ 2^y = 694 + 1387x \tag{1} $$

WolframAlpha gives the following solutions:

$$x = 94, y = 17$$

$$x = 24772702, y = 35$$

$$x = 6494015324254, y = 53$$

$$x = 1702367153161371742, y = 71$$

$$x = 446265334998334634066014, y = 89$$

If we observe the $y$ coordinates, we see a pattern:

$$17 = 17 \times 1 + 0$$

$$35 = 17 \times 2 + 1$$

$$53 = 17 \times 3 + 2$$

$$71 = 17 \times 4 + 3$$

$$89 = 17 \times 5 + 4$$

Now consider the Exponential Diophantine equation:

$$ 3^y = 925 + 1387x \tag{2} $$

Again, from WolframAlpha, we get the following solutions:

$$x = 36071770078586, y = 35$$

$$x = 5414179174462671091526111000906, y = 71$$

The pattern for the $y$ coordinates is

$$35 = 35 \times 1 + 0$$

$$71 = 35 \times 2 + 1$$

We did not construct the Diophantine equations (1) and (2) at random. In fact, they were carefully constructed by solving the equations

$$ 2u - 1387v = 1 \tag{3} $$

and

$$ 3p - 1387q = 1 \tag{4} $$

respectively. $u = 694$ is a particular solution of Eqn. (3) and $p = 925$ is a particular solution of Eqn. (4).

Questions:

1. Can we generalize that for an Exponential Diophantine equation of the form

$$g^y = ax + b \tag{5}$$

where $\gcd(g,a) = 1$ and $u = b$ is a particular solution to $gu - av = 1$ and if $y = y_0$ is a particular solution of Eqn. (5) then $y = 2y_0 + 1$ is also a solution of Eqn. (5)?

2. If the answer to Question 1 is affirmative, then can we use this fact to actually solve equations of the type Eqn. (5)?

Note: This question is related to this other MSE question I had asked a few days ago.

Answer 1

The general solution can be found as follows.

Let $(x_0,y_0)$ be any solution of $g^y=ax+b$ and let $X,Y$ be the solution of $g^y=ax+1$ with $y$ as small a positive integer.

Then all the solutions for $y$ are given by $$y=y_0+nY.$$

NOTE

It is easy to see that such a $y$ will always give a solution because then $$g^y=g^{y_0}(g^Y)^n=(ax_0+b)(aX+1)^n$$ where the RHS is clearly a multiple of $a$ plus $b$, as required.

EXAMPLE

In your first example you correctly obtained the formula $17(n+1)+n$.

Your answer can be put into the same form as the general formula because it can be written as $17+18n$.