India — Team Selection Test, 2012

Observe $b^2 - 1 = b(a+b) - (ab+1)$ and $b^2 - 1 = b(a-b) - (ab-1)$. Hence $a+b$ and $a-b$ both divide $b^2 - 1$. Thus $\text{lcm}(a+b, a-b)$ divides $b^2 - 1$. Since both are positive, $\text{lcm}(a+b, a-b) \le b^2 - 1$.
Let $d = \text{gcd}(a, b)$. Then $d|ab$ and $d|a+b|ab+1$. Hence $d|1$ showing $d=1$. If $e = \text{gcd}(a+b, a-b)$, then $e|2a$ and $e|2b$ so that $e|\text{gcd}(2a, 2b)$. But $\text{gcd}(2a, 2b) = 2\ \text{gcd}(a, b) = 2$. Thus $e|2$ and hence $e \le 2$. Hence
$$\text{lcm}(a+b, a-b) = \frac{(a+b)(a-b)}{\text{gcd}(a+b, a-b)} \ge \frac{a^2-b^2}{2}.$$
It follows that $a^2 - b^2 \le 2(b^2 - 1)$ or $a^2 - 3b^2 \le -2$.
Suppose $a$ and $b$ are positive integers such that $a^2 - 3b^2 = -2$. Then $a$ and $b$ have same parity. For such a pair $(a, b)$, we have
$$ab + 1 = ab + \frac{3b^2 - a^2}{2} = (a+b)\frac{3b-a}{2},$$
$$ab - 1 = ab - \frac{3b^2 - a^2}{2} = (a-b)\frac{a+3b}{2}.$$
Hence $a+b$ divides $ab+1$ and $a-b$ divides $ab-1$. We also observe that
$$\sqrt{3}b = \sqrt{a^2 + 2} < \sqrt{a^2 + 2a + 1} = a + 1,$$
so that $a > \sqrt{3}b - 1$.
Thus we look for solutions of the equation $x^2 - 3y^2 = -2$ in positive integers. This equation has infinitely many solutions which may be described as follows:
The equation $x^2 - 3y^2 = -2$ has a particular solution $(1, 1)$. Consider the equation $x^2 - 3y^2 = 1$. This has infinitely many solutions $(u_n, v_n)$ given by
$$u_n + \sqrt{3}v_n = (2 + \sqrt{3})^n.$$
Let $a_n = u_n + 3v_n$ and $b_n = u_n + v_n$. Then
$$a_n^2 - 3v_n^2 = -2(u_n^2 - 3v_n^2) = -2.$$
For $n \ge 1$ we have $b_n > 1$.