$ ab|a^2+b^2+k$ Let $ a_0=1,b_0|k+1$ and and Let $ k+1=mb_0$ $ d=\frac{mb+b^2}{b}=m+b_0$ Consider the equation: $ a^2+b^2-dab+k=0$ $ \Delta=(d^2-4)b^2-4k$ Chose : $ (d^2-a)b^2-4k=y^2$ $ \Leftrightarrow y^2-(d^2-4)b^2=-4k$ This equation has an solution $ (y,b)$ so it has infinite solution.

Tell me if it needs translate http://www.irysc.com/forum/t10301/#post126935 I construct a sequence that satisfy $a_{n+2} . a_n = {a_{n+1}}^2 + k$. From this relation we conclude that \[ a_{n}\;\text{divides}\; a_{n+1}^{2}+k\;\;\text{and}\;\; a_{n+1}\;\text{divides}\; a_{n}^{2}+k \]

Define the sequence $a_{n}$ such that $a_{1} = 1,a_{2} = k+1,a_{n+2} = (k+2)a_{n+1}-a_{n}.$ And we can prove the following equation by induction. $(a_{n+1})^{2}+(a_{n})^{2}-(k+2) a_{n+1}a_{n} = 0.$ So,we have done.