i smel here pell-fermat. the soultion of $x^2-2y^2=1$ is $x_{n}+\sqrt{2}y_{n}=(3+2\sqrt{2})^{n}$. .......

Too bad I don't smell it... have you read the problem correctly?

2xy > x+y if x,y >= 2 So x = 1 The equation then reduces to y(y-2) = 0 So (x,y) = (1,2) only.

markmichell wrote: 2xy > x+y if x,y >= 2 And what's from it? I think you misunderstood the problem, didn't you?

its not so hard.since $x,y$ are positive integers and the equation can be writen as $(x+y)^{2}-(xy)^{2}=1+(xy)^{2}>0$ it implies that $x+y-xy>0$ so $0\leq y<E(\frac{x}{x-1})$ ($E()$ est la partie entiere de ) so iff $x=2$ then the value of $y$ are 0.1.2 and if $x=1$ we return to the eqution $y=0,2$ and iff not the only value of $y$ are 0,1 so solutions are $(0,1).(1,0).(2,1).(1,2)$.

Generator00 wrote: its not so hard.since $ x,y$ are positive integers and the equation can be writen as $ (x + y)^{2} - (xy)^{2} = 1 + (xy)^{2} > 0$ it implies that $ x + y - xy > 0$ so $ 0\leq y < E(\frac {x}{x - 1})$ ($ E(?)$ est la partie entiere de ?) so iff $ x = 2$ then the value of $ y$ are 0.1.2 and if $ x = 1$ we return to the eqution $ y = 0,2$ and iff not the only value of $ y$ are 0,1 so solutions are $ (0,1).(1,0).(2,1).(1,2)$. x,y is must be positive solution. I think (2,1),(1,2) is only

because (x+y)^2<=2(x^2+y^2) (x+y)^2-2(xy)^2=1<=2(x^2+y^2)-2(xy)^2 2(x^2-1)(y^2-1)<=1 if x,y>=2 2(x^2-1)(y^2-1)>=18>1 contradiction x<2 or y<2 wlog x<2 x=1(x>0) y=0or 2 y=2(y>0) and x=2 y=1 solution is (1,2) , (2,1)

very easy Pell equations

Abror wrote: very easy Pell equations It's not Pell's equation!!!! Pell's equation is $x^2-dy^2=1$ where it's not necessary for $x$ and $y$ to be in the form of $a+b$ and $ab$, respectively. This just looks like Pell's equation but there is a limitation.