If we consider quadratic equation wrt $ y $ we have $ 5x^2-d^2=1 $ ( where $ d $ is discriminant of the q-equation ). It's Pell equation.

$$x(x+y)=y^2+1$$$$\Rightarrow y=\frac{x\pm\sqrt{5x^2+4}}{2}$$$$\Rightarrow y=\frac{x+\sqrt{5x^2+4}}{2}$$$$\Rightarrow 5x^2+4\text{ is a perfect square}$$$$\Rightarrow 5x^2+4=p^2$$If $x$ is even$:$ $$\Rightarrow x=2k$$$$\Rightarrow 20k^2+4=p^2$$$$\Rightarrow t \text{ is even}$$$$\Rightarrow p=2q$$$$\Rightarrow 20k^2+4=4q^2$$$$\Rightarrow q^2-5k^2=1 \text{ is Pell's equation}$$If $x$ is odd$: \text{ need to finish}$

hectorleo123 wrote: $$(x,y)=(1,1)\text{ is the only solution}_\blacksquare$$ What about $(x,y)=(5,8)$ ?

pco wrote: hectorleo123 wrote: $$(x,y)=(1,1)\text{ is the only solution}_\blacksquare$$ What about $(x,y)=(5,8)$ ? Can you help me with my solution please? @below Thanks

hectorleo123 wrote: Can you help me with my solution please? As Sardor said, this is just Pell equation : $(2x+y)^2-5y^2=4$