The equation is like this: $3^x -2^y = 19^z$
It seems that no way to find the solution except using trial and error.
I got only one solution:
$x=3, y=3, and z= 1$ by using trial and error.
But, when I use the software, I got the bad value:
http://www.wolframalpha.com/input/?i=3^x+-2^y+%3D+19^z
Is there any elegant way to get the solution(s) ?
Thanks
If $x$ or $z$ is even, there is one and only one solution, namely, $x=2, y=3, z=0$, if $x$ is odd and is divisible by $3$, there is one and only one solution namely, $x=3, y=3, z=1$
given; $3^x-2^y=19^z$.........$eqn 1$ we have, $-2^y \equiv 1 \pmod 3$ or $2^y \equiv 2 \pmod 3$ or $2^{y-1} \equiv 1 \pmod 3$ hence $y-1 \equiv 0 \pmod 2$ by Euler-Fermat theorem. So, y must be odd. If $y>1$ Eqn1 becomes $(-1)^x \equiv (-1)^z \pmod 4$, so that $y$ and $z$ have the same parity. If $x>1$, eqn1 becomes $-2^y \equiv 1 \pmod 9$ and hence, by Euler-Fermat theorem $y \equiv 0 \pmod 3$. $x, y, z>1$, let $x$ be even, then eqn1 becomes $(-1)^k-2^y \equiv 1 \pmod 5$ for $x=2k$, if $k$ is even, the given equation has no solution, if $k$ is odd, i.e $x \equiv 2 \pmod 4$ then $2^y \equiv 2 \pmod 5$ and by Euler-Fermat (E-F), $y \equiv 1 \pmod 4$, similarly, if $x$ is odd, there is no solution if $x \equiv 1 \pmod 4$. Working $\pmod 8$ and using Euler-Fermat, we find, if $z$ is even, then $z \equiv 0 \pmod 4$, if $z$ is odd then, $z \equiv 1 \pmod 4$. Now if $x=2m$, then $z=2n$ and the given equation is $3^{2m}-19^{2n}=2^y$ or $(3^m-19^n)(3^m+19^n)=2^y$, we must have $(3^m-19^n)=2^r$ and $(3^m+19^n)=2^{y-r}$ where $3^m=2^{r-1}+2^{y-r-1}$ and $19^n=-2^{r-1}+2^{y-r-1}$ from which we must have $r-1=0$ since $r<y-r$. The last equation becomes $19^n=-1+2^{y-2}$, now, since $4|z$ it follows that $2|n$, and hence, $19^n$ is a perfect square, hence $19^n \equiv 1 or 0 \pmod 4$ but for $y>3$, the last equation becomes $19^n \equiv -1 \pmod 4$ a contradiction. Hence, if $x$ or $z$ is even and $y>3$, there is no solution, for $y \leq 3$ solutions can be checked manually. We now turn to the case where $x$ is odd. If $x$ is odd and divisible by $3$, then since $3|y$, the left hand side of the given equation is factorizable and by considering the factors, following a method similar to the above, $\pmod 4$, shows there is no solution except the case presented above....NOTE; if $3|x$ then $3$ does not divide $z$ otherwise, since $3|y$, by Fermat Last Theorem there is no solution. If $3$ does not divide $x$ it may divide $z$ and similar to what is shown above, the analysis can be carried a step further by considering the factors of the right hand side of the equation $3^x=2^y+19^z$, in which case one finds, working in $\pmod 4$, there is no solution.