Can this be accepted as a proof of the question.

Let $a,bin+mathbb Z$ 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 $pin +mathbb Z$ and $n,minmathbb Q.$

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 $pin +mathbb{Z}$ and $m,nin mathbb{Q}.$

Can this be a proof?