Peter wrote: Determine all pairs $ (a, b)$ of integers for which $ a^{2} + b^{2} + 3$ is divisible by $ ab$. if (a,b) is solution, then $ (\pm a, \pm b) and ( \pm b, \pm a)$ are solutions. $ a,b\neq 0$, then we consider the solutions (a,b) in the positive integers, such that $ a \ge b$ if (x,y) is solution, with x>y>1, then $ txy=x^2+y^2+3>x^2$ ,then ty>x. $ x(ty-x)=y^2+3 \le (x-1)^2+3=x^2-2x+4<x^2$ ,then ty-x<x. (ty-x,y) or (y,ty-x) is solution and ty-x + y<x+y. each solution with x>y>1, through this process to creat a end solution, (r,s); $ r \ge s$, such that r=s,or s=1, because the sume of components decreases if r=s, $ r^2$ divides $ 2r^2+3$, then $ r^2$ divides 3, then r=1 if s=1, r divides $ r^2+4$ ,then r divides 4, then r=1,2,4. the possible end solution are: (1,1), (2,1), (4,1). the "t" respective are 5,4,5 hence the solutions are $ (x_n,y_n) and (z_n,w_n)$ $ x_0=1, y_0=1, x_{n+1}=5x_n -y_n, y_{n+1}=x_n$ (t=5) $ z_0=2, w_n=1, z_{n+1}=4x_n -y_n, w_{n+1}=z_n$ (t=4)

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