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

  • $\begingroup$ Taking log both sides may help? Just an idea. $\endgroup$ Apr 30, 2015 at 7:24

2 Answers 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.

  • $\begingroup$ I don't understand how equation $2^x+7^y=19^z$ gives $2^x+1 \equiv 1 \;(\bmod\; 6)$ ? $\endgroup$
    – vito
    Apr 30, 2015 at 8:46
  • 2
    $\begingroup$ @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 $a+c\equiv b+d\mod m$. $\endgroup$
    – Apurv
    Apr 30, 2015 at 12:04

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}$ .


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