Indija TS 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 $b_n>1$.