Solve this equation $12^x-5^y=19$ positive integers Find all $x,y$ be positive integers,such
$$12^x-5^y=19$$
I found $(x,y)=(2,3)$ is solution,maybe have other,so I consider case  $x,y>3$ and $\pmod 9$,since
$$12^x\equiv 0\pmod 9,x\ge 2$$
then $5^y\equiv -1\pmod 9$,
$5^1\equiv 5 \pmod 9$
$5^2\equiv 7 \pmod 9$
$5^3\equiv 8 \pmod 9$
$5^4\equiv 4 \pmod 9$
$5^5\equiv 2 \pmod 9$
$5^6\equiv 1 \mod 9\Rightarrow y\equiv 3 \mod 6\Rightarrow y$ is odd.
 A: To show that there are no solutions with $x \ge 3$, let's consider the equation modulo $27$. When $x\ge 3$, the equation becomes $-5^y \equiv 19 \pmod{27}$, or $5^y \equiv 8 \pmod{27}$. This calculation is intimidating, but rewarding: $5$ has multiplicative order $18$ modulo $27$, and so we are able to conclude $y \equiv 15 \pmod{18}$. In particular, $y \equiv 6 \pmod 9$.
To make use of this, we look at the factors of $5^9-1$: two reasonably-sized ones are $19$ and $31$. We want factors of $5^9-1$ because when $d \mid 5^9-1$, knowing that $y \equiv 6 \pmod 9$ tells us that $5^y \equiv 5^6 \pmod d$.

*

*Taking the equation mod $19$, we get $12^x - 5^6 \equiv 0 \pmod{19}$, or $12^x \equiv 7 \pmod{19}$. This is true when $x \equiv 4 \pmod 6$.

*Taking the equation mod $31$, we get $12^x - 5^6 \equiv 19 \pmod{31}$, or $12^x \equiv 20 \pmod{31}$. This is true when  $x \equiv 2 \pmod {30}$.

But these two contradict each other, so there are no solutions (with the assumption $x\ge3$ we started from).
We could also have combined the last two steps into a calculation modulo $19\cdot31 = 589$, where $12^x - 5^6 \equiv 19 \pmod{589}$ has no solutions.
