An exponential Diophantine equation.

Question

I'm solving an exponential Diophantine equation, and I've taken (mod 3) to have that $z$ is odd, but it's still hard for me .

" Find all the positive integers $x, y, z$ such that $7^x+1=3^y+5^z$ ".

Answer 1

Finally, I've found a solution only using modulo.

We want to find all the positive integers $x,y,z$ such that

$$ 7^x+1=3^y+5^z \tag 1 $$ Consider the equation $(1)$ in $\pmod 3$, we have $$ 5^z \equiv 2\pmod 3$$ So $z$ is odd. If $y\ge 2$, then we can take $\pmod 9$ to have that $$ 7^x+1\equiv 5^z \pmod 9$$ Note that $ord_9(5)=6$ and $ord_9(7)=3$, we have:

If $$x\equiv 1 \pmod 3 \Rightarrow z\equiv 3 \pmod 6$$ $$x\equiv 2 \pmod 3 \Rightarrow z\equiv 1 \pmod 6$$ $$x\equiv 0 \pmod 3 \Rightarrow z\equiv 5 \pmod 6$$

Now take $(1)$ in $\pmod 7$, we have $$3^y+5^z\equiv 1 \pmod 7$$ So if $$z\equiv 1 \pmod 6 \Rightarrow y\equiv 1 \pmod 6$$ $$z\equiv 3 \pmod 6\Rightarrow y\equiv 2 \pmod 6$$ $$z\equiv 5 \pmod 6 \Rightarrow y\equiv 5 \pmod 6$$

Then take $\pmod {13}$ we have $$7^x+1\equiv 3^y+5^z\pmod {13}$$ Now consider all cases of $x$ in modulo 12, we found only one case is compatible: $x\equiv 2 \pmod 12$, $ y\equiv 1 \pmod 6$, and $z\equiv 7 \pmod 12$.

Now we can write $x=12k+2$, then $7^x+1=49^{6k+1}+1\equiv 0 \pmod 5\Rightarrow 3^y+5^z\equiv 0 \pmod 5$, contradiction.

So we conclude that $y=1$ and $7^x-5^z=2$. If $z\ge 2$ then take modulo 25 to have $7^x\equiv 2 \pmod {25}$, contradiction. So $z=1$ and $x=1$.

Hence $(x,y,z)=(1,1,1)$ is the only solution for $(1)$.