<s>[INST] Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ To find the triples, I had some tries.
First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.  


*

*case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.

*case 2. One of them are even, the other is odd. Then $y^2+z^2+1=2\pmod 4$.

*case 3. Both $y$ and $z$ are odd. Then $y^2+z^2+1=3\pmod 4$.  


Thus for $x\ge 2$, there is no solution. Then $x=0$ or $x=1$.
Now for $x=0$, $2018^0=y^2+z^2+1$ ,then $y^2+z^2=0$, hence we have $y=0$ and $z=0$.
For $x=1$, $2018^1=y^2+z^2+1$, then my question is how to solve $y^2+z^2=2017$.
Also since I didn't learn Number Theory systematically, I don't know if I was proceeding in the right way for this question.
Lastly, Are there more general methods for problems like this?
 [/INST] You have solved almost all cases. So let me add the solution for $x=1$, and exclude negative $x$.
Since $2017$ is a prime which is congruent $1$ modulo $4$, it is representable as the sum of two squares. The solutions are, up to sign and order, given by
$$
2017=44^2+9^2,
$$
see here:
$2017$ as the sum of two squares
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