$2^x+7^y=19^z$ has no solution in positive integers $x$, $y$, $z$

Question

How do I show that the diophantine equation $2^x+7^y=19^z$ has no solution in positive integers $x$, $y$, $z$

Answer 1

Hint: try finding solutions modulo n for some n.

Edit: As pointed out in the comments this is the obvious thing to do, but I'm not sure if you, the OP, are aware of that. A small value of n suffices and so just trying different values of n will get you somewhere. However if you want to be clever you'll notice there's a value of n with $ 7^y \equiv 1 \mod{n}$ and $ 19^z \equiv 1 \mod{n}$ .

Answer 2

Going modulo $6$, the equation gives $$2^x+1\equiv 1 \pmod 6$$So, $$2^x\equiv 0 \pmod 6$$But no power of $2$ is divisible by $6$.

Which shows that no solutions exist. Hence proved.