Solutions in positive integers of $a+b\mid ab+1$ and $a-b\mid ab-1$ I am interested in a proof for the following claim. Suppose that for integers $a>b>1$ the following two conditions hold:
$$a+b\mid ab+1,$$ $$a-b\mid ab-1.$$
Then $\frac{a}{b}<\sqrt{3}$. Furthermore, is it possible to determine all positive integer solutions in this case ? 

To prove the bound, note first that necessarily $\gcd (a,b) = 1$, and then write
$$\begin{align}ab+1 &= (b-\gamma)(a+b)\\
\iff \gamma(a+b) &= b(a+b) - ab - 1 = b^2 -1\\
ab - 1 &= (b+\delta)(a-b)\\
\iff \delta(a-b) &= ab - 1 - b(a-b) = b^2-1.
\end{align}$$
So both, $a+b$ and $a-b$ divide $b^2-1$. Since $\gcd (a+b,a-b) = \gcd(a+b,2) \in \{1,\,2\}$, we have
$$\frac{(a+b)(a-b)}{\gcd(a+b,a-b)} \mid b^2-1 \Rightarrow \frac{a^2-b^2}{2} \leqslant b^2-1 \Rightarrow a^2 \leqslant 3b^2 - 2.$$
