Quadratic Diophantine equation

Question

how one can approach solving diophantine equation $a^{2}$+ $b^{2}$ - $1$ = $n {\cdot}a{\cdot}b$ in positive integers $a$, $b$, $n$ ( where also $a$ and $b$ are relatively prime)? I have tried so far only to treat it as quadratic one, and solve for discriminant, but am unable to get decent result. Also, i got $a{\cdot}b$|($a-b-1){\cdot}(a-b+1)$, and am not sure where to move from that result. Thanks in each case, any help or resource for solving these is appreciated.

Answer 1

We demand $n \geq 3.$

as you have $x^2 - nxy+y^2 = 1,$ all solutions come from applying the mapping $$ (x,y) \mapsto (nx-y , x ) $$ beginning with the pair $(1,0).$

Along with those, one may switch $(x,y) $ to $(y,x).$ Also negate both $(-x,-y)$

$$ (1,0); \; (n,1); \; (n^2-1,n); \; (n^2-1,n); \; (n^3 - 2n, n^2 - 1 ); ........ $$