Find all positive integer pairs $(a, b)$ such that $$(ab + a + b) \mid (a^2 + b^2 + 1)$$
Let $a^2 + b^2 + 1 = k(ab + a + b), k \in \mathbb N, k \ge 1$.
For $k = 1$, we have that $$a^2 + b^2 - ab - a - b + 1 = 0$$, where $(a, b) = (1, 1)$ satisfies the above condition.
For $k = 2$, we have that $$a^2 + b^2 - 2ab - 2a - 2b + 1 = 0$$, where $(a, b) = (n^2, (n + 1)^2)$ where $n \in \mathbb N$ satisfies the condition, since $$(a - b)^2 - 2(a + b) + 1 = [(n + 1)^2 - n^2]^2 - 2[(n + 1)^2 + n^2] + 1$$
$$ = (2n + 1)^2 - 2(2n^2 + 2n + 1) + 1 = (4n^2 + 4n + 1) - (4n^2 + 4n + 2) + 1 = 0$$
I don't know what to do next for the case of $k \ge 3$. Any help would be greatly appreciated.
Reference is HURWITZ 1907
Caution: this is about POSITIVE $x,y$ values. For $k=5,$ there is a solution at $-5,-3.$
The main thing is that the jumping step leads to inequalities for what Hurwitz called ground solutions. We have $$ x^2 - kxy + y^2 -kx-ky +1=0 $$ To jump, the coefficient of $x$ is $-ky-k.$ The two $x$ solutions with $y$ fixed are $x$ and $x' = ky+k - x.$ A Grundlosung (it is an o umlaut) occurs when $x' \geq x$ and $y' \geq y.$ In turn, this says $$ 2x \leq k(y+1) \; \; , \; \; \; \; 2y \leq k(x+1) $$ In the diagrams below, this is the region of the hyperbola in the first quadrant between the two lines. As you can see, there are actually no integer points on the arc between the two lines. For example, intersecting the hyperbola with $y = \frac{k}{2}(x+1)$ leads to $$ x = \frac{6 k^2 \pm \sqrt{32} \sqrt{k^4 + k^2 - 2}}{4 - k^2} \; \; \; . \; \; $$ As soon as $k \geq 3,$ we find $$ \sqrt{k^4 + k^2 - 2} < k^2 + \frac{1}{2} $$ so that the numerator is positive for either $\pm,$ so that the final $x$ values are both negative. In turn, this means that there are NO integer points along the arc between the Hurwitz lines. There are, in fact, no fundamental solutions, therefore no (positive) integer solutions when $k \geq 3.$