Exponential equation in $\mathbb{Z}$ Solve the equation:
$$5^{18x}+12=13^{18y+1},$$ where x, y are positive integer.
I notice that $x=y=0$ is the only solution. I tried to reduce modulo primes but I didn't succeed.
 A: Suppose $x \neq 0$. Modulo $5$ we find that
$$2 \equiv 3^{18y+1} \equiv 3 \cdot 4^y\hspace{10pt} \mbox{ mod } 5$$
since by Euler-Fermat's theorem the exponent can be reduced modulo $\varphi(5)=4$. So looking at the powers of $4$ mod $5$ ($4^2 \equiv 1$) we conclude that $y$ must be odd.
Now, modulo 13 we have
$$5^{18x} \equiv 1 \hspace{10pt} \mbox{ mod } 13$$
and therefore if we examine the powers of $5$ modulo 13 ($5^2 \equiv -1$, $5^3 \equiv -5$, $5^4 \equiv 1$) we obtain
$$18x \equiv 0 \hspace{10pt} \mbox{ mod } 4$$
and so $x$ must be even.
Finally if we look at the equation modulo $37$ substituting $x=2k$ and $y=2h+1$ we get
$$5^{36k}+12 \equiv 13^{36h+19} \hspace{10pt} \mbox{ mod } 37$$
and hence, since $\varphi(37)=36$ and $13$ has an inverse mod $37$
$$1+12=13 \equiv 13^{19} \Rightarrow 13^{18} \equiv 1  \hspace{10pt} \mbox{ mod } 37$$
But $$13^{18} \equiv 21^9 \equiv 11^3 \equiv 110 \equiv -1 \hspace{10pt} \mbox{ mod } 37$$
so this is absurd. Therefore we cannot have $x \neq 0$, and if $x=0$ then immediately $y=0$.
A: $$5^{18x}+12=13^{18y+1}\Leftrightarrow 10^{9x}+12=13^{18y+1}$$
Assume $x\neq 0$, modulo 10, we have
$$2\equiv 3^{18y+1}\pmod{10}$$
Due to $3^{\varphi(10)}=3^{4}\equiv1\pmod{10}$, then
$$2\equiv3^{2y+1}\pmod{10}$$
According CRT(Chinese remainder theorem), we have
\begin{equation}
\begin{cases}
2\equiv 3^{2y+1}\pmod{2} \\
2\equiv 3^{2y+1}\pmod{5} \\
\end{cases}
\end{equation}
However,
$$
2\equiv3^{2y+1}\pmod{2}\Leftrightarrow 0\equiv 1\pmod{2}
$$
Which is impossible, so we are sure that $x=0$, then $y=0$.
