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

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

• Taking log both sides may help? Just an idea. – Kushashwa Ravi Shrimali Apr 30 '15 at 7:24

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 $6$.
• I don't understand how equation $2^x+7^y=19^z$ gives $2^x+1 \equiv 1 \;(\bmod\; 6)$ ? – vito Apr 30 '15 at 8:46
• @vito, $7\equiv 1\mod 6\implies 7^y\equiv 1\mod 6$ and $19\equiv 1\mod 6\implies 19^z\equiv 1\mod 6$. Also if $a\equiv b\mod m$ and $c\equiv d\mod m$, then $a+c\equiv b+d\mod m$. – Apurv Apr 30 '15 at 12:04
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 $19^z \equiv 1 \mod{n}$ .