If $\log_3(A)$ is rational, can $\log_3(A+1)$ be rational? (Where $A$ is a positive real number.) Question:
If $\log_3(A)$ is rational, can $\log_3(A+1)$ be rational? (Where $A$ is a positive real number.)

My Working:
Starting by assuming
$$\log_3(A) = x \ {\rm{and}} \ \log_3(A+1) = y$$
it gives $$3^x = A \ {\rm {and}} \ 3^y = A + 1$$
Which further implies that,
$$3^y = 3^x + 1$$
and
$$3^y - 3^x = 1$$
$$3^{x}(3^{y-x}- 1) = 1$$
$$3^x = 1\ {\mathrm {and} }\ 3^{y-x}  = 2 $$
$$x = 0\ {\mathrm {and} }\ 3^{y}  = 2 $$
Which further gives us:
$$y = \frac{\log(2)}{\log(3)}$$
and hence $y$ is irrational.

Is my work correct?
 A: More generally, let's determine when $\log_3(A)$ and $\log_3(A+m)$ are both rational for any positive integer $m$. Set $x = \log_3(A)$ and $y=\log_3(A+m)$, so that $3^y = 3^x + m$. Suppose that $x=\frac ab$ and $y=\frac cd$ are rational (in lowest terms).

*

*Suppose first that $b>d$. Then $3^c = (3^x+m)^d$, and so $3^x$ is a root of the polynomial $(T+m)^d - 3^c$ which has integer coefficients. But this has smaller degree than the minimal polynomial of $3^x = \sqrt[b]{3^a}$, namely $T^b - 3^a$, which is a contradiction.

*Similarly, suppose that $d>b$. Then $3^a = (3^y-m)^b$, and so $3^y$ is a root of $(T-m)^b + 3^a$; but this has smaller degree than the minimal polynomial of $3^y = \sqrt[d]{3^c}$, which is $T^d-3^c$.

*Finally, suppose that $b=d$. Then $3^x$ again has minimal polynomial  $T^b - 3^a$ but also is a root of the polynomial $(T+m)^b - 3^c$. Since these are monic of the same degree, we must have $T^b - 3^a = (T+m)^b - 3^c$. This is impossible if $b>1$ (look at the coefficient of $T^k$ for any $0<k<b$), and therefore it's necessary that $b=d=1$, so that $x$ and $y$ are actually integers.

In conclusion, we can only have $\log_3(A)$ and $\log_3(A+m)$ simultaneously rational if both $A$ and $A+m$ are integer powers of $3$. In particular, this doesn't happen when $m=1$.
