Let $a,b\in+\mathbb Z$ such that $\dfrac{a^2+b^2}{ab+1}=k.$
Prove that $k$ is the square of an integer when $$a^2+b^2$$ is divisible by $ab+1$.
Soln. For $k$ to be a perfect square, $\dfrac{a^2 + b^2}{1+ab}$ must be in the form $\dfrac{mp^{n+2}}{mp^n}$ where $p\in +\mathbb Z$ and $n,m\in\mathbb Q.$
Hence $\dfrac{a^2+b^2}{ab+1} = \dfrac{mp^{n+2}}{mp^n}.$ On comparison of numerator and denominator we have $$ a^2 + b^2 = mp^{n+1} \tag 1 $$ and $$ ab+1 = mp^n. \tag 2 $$ Solving for $a$ and $b$ we get \begin{align} b & =\sqrt{\frac{mp^{n+2} \pm \sqrt{m^2 p^{2n+4} -4m^2 p^{2n} +8mp^n -4}}{2}} \\[6pt] \text{and } a & = (mp^n -1)\sqrt{\frac{2}{mp^{n+2} \pm \sqrt{m^2 p^{2n+4} -4m^2 p^{2n} +8mp^n -4}}} \end{align} so $k$ will be perfect square when $a$ and $b$ can be expressed in this form where $p\in +\mathbb{Z}$ and $m,n\in \mathbb{Q}.$
Can this be a proof?