The Diophantine equation X⁴-Y⁴-X=0

Let’s solve the Diophantine equation X⁴-Y⁴-X=0 together so as to see a standard approach that is encountered in solving Diophantine equations of the form Z(f(Z)-1)=Y² since f is a function of Z. For any typos I invite you to report them in the comments. Let’s suppose that (X,Y) is a non-trivial solution, that is, X is different from 0. Then X⁴-X=Y⁴ or X(X³-1)=Y⁴. We have two possibilities: X>0 or X0 then X³-1>0 but (X,X³-1)=1 because if d|X and d|X³-1 then d|X³-(X³-1) that is, d|1 and therefore d=1 therefore there must exist (see the handouts on Fermat’s last theorem) two natural numbers a and b coprime such that X=a² and X³-1=b². Then a⁶-b²=1 and that is (a³+b)(a³-b)=1 from which a = 1 and b=0. Therefore X=1 and Y=0. If instead X<0 then X³-1<0 therefore -X(1-X³)=Y⁴ implies the existence of two natural numbers c and d coprime such that -X=c² and 1-X³=d² from which d²=1-(-c²)³=1+c⁶ and therefore d²-c⁶=1 follows that c=0 and d=1 and therefore X=0 is absurd. Therefore the only solutions are (0,0) and (1,0).