Case $1$: One of $x,y$ is $0$. Note that $y=0$ implies $x=0$ and $x=0$ implies $y=0$ hence this case is done, producing solution $(0,0)$ Case $2$: $x,y$ are in $N$ Let $x=ky$ where $k$ is rational, so $k$ is strictly positive. We get that $k^{2018}-1=y^{2016}k^{2017}$ upon cancellation. Now if $k^{2018}-y^{2016}k^{2017}-1=0$ is irreducible in $Q$, the set of rationals, it is also irreducible in $Z$, the set of integers, because the leading coefficient is $1$. So if $k^{2018}-y^{2016}k^{2017}=1$ has a rational solution, it also has an integer solution. Let the integer solution be $z$. Thus $z^{2018}-y^{2016}z^{2017}=1$ or $(z^{2017})(z-y^{2016})=1$. Both are integers, and $z^{2017}$ is definitely positive, thus $z^{2017}=z-y^{2016}=1$. But this means $z=1,y=0$ thus $x=0$, leading back to the same solution, thus $(0,0)$ is the only solution. Proved.

is there any other ways to solve this?

adorefunctionalequation wrote: is there any other ways to solve this? Suppose there exist a prime $p \mid y$ (otherwise $y=1$ and there are no solutions). By definition, $p \mid (xy)^{2017} =x^{2018}- y^{2018} \implies p \mid x$, this means that $p \mid y$ if and only if $p \mid x$. WOLG suppose $v_p(x) > v_p(y) \implies 2018v_p(y)=v_p(y^{2018}) = v_p(x^{2018}- y^{2018}) = v_p((xy)^{2017})=2017(v_p(x)+v_p(y)) \implies v_p(y)=2017v_p(x)$ and this is a contradiction. So $v_p(x)=v_p(y)$ for all primes $p \mid x,y \implies x=y$. Substituting in the problem's equation: $x^{2018}- x^{2018}=x^{4034}=0$. Therefore $x=y=0$ is the only solution.

HyperDunteR wrote: adorefunctionalequation wrote: is there any other ways to solve this? Suppose there exist a prime $p \mid y$ (otherwise $y=1$ and there are no solutions). By definition, $p \mid (xy)^{2017} =x^{2018}- y^{2018} \implies p \mid x$, this means that $p \mid y$ if and only if $p \mid x$. WOLG suppose $v_p(x) > v_p(y) \implies 2018v_p(y)=v_p(y^{2018}) = v_p(x^{2018}- y^{2018}) = v_p((xy)^{2017})=2017(v_p(x)+v_p(y)) \implies v_p(y)=2017v_p(x)$ and this is a contradiction. So $v_p(x)=v_p(y)$ for all primes $p \mid x,y \implies x=y$. Substituting in the problem's equation: $x^{2018}- x^{2018}=x^{4034}=0$. Therefore $x=y=0$ is the only solution. Nice Solution Thanks

if $x = 0 $ then $y = 0$ if $ y=0$ then $x=0$ if $x=y$ then $x=y=0$ so now suppose we have $x>y$ since RHS should be non negative then let $x = dm ,y =dn$ where $\gcd(m,n) = 1 ,\gcd(x,y) = d$ $m^{2018}-n^{2018} = d^{2016}m^{2017}*n^{2017}$ \mod m $n^{2018} \equiv 0 \mod m$ but $\gcd(m,n) = 1$ this lead to $m = 1$ \mod n $m^{2018} \equiv 0 \mod n$ with similar reasoning $n = 1$ so $0 = d^{2016}$ which show that $d = 0$ so the only solution is when $x= y= 0$