Indija TS 2008

Let $a$, $b$, $c$ be a solution of the given equation. Putting $$\frac{a}{b}=X,\frac{b}{c}=Y,$$ we have $c/a=(1/XY)$. Note that $X$ and $Y$ are rational numbers. The equation takes the form $$X^{2}Y+X{Y}^2+1=3XY.(1)$$ This is a cubic curve. We observe that $(1,1)$ is a point on the curve (1). If we know the rational points on (1), then we can get integer solutions of the given equation. Suppose $(x,y)$ is a rational point on the curve (1). Then the line joining $(x,y)$ to $(1,1)$ has rational slope. Conversely, any line through $(1,1)$ having rational slope intersects the curve (1) in a point having rational coordinates. Consider the line $$\frac{y-1}{x-1}=m,$$ where $m\ne -1$ is a non-zero rational number. Let its intersection with (1) be $(X_0,Y_0)\ne (1,1)$. Then we have $Y_0=m(X_0-1)+1$. Putting this in (1), we get $$X_0^2(m(X_0-1)+1)+X_0(m(X_0-1)+1{)}^2+1=3X_0(m(X_0-1)+1).$$ This reduces to $$(X_0-1{)}^2(m(1+m)X_0+1)=0.$$ Since $X_0\ne 1$, we have $$X_0=-\frac{1}{m(1+m)},Y_0=m(X_0-1)+1=-\frac{m^2}{(1+m)}.$$ Thus $$\frac{a}{b}=-\frac{1}{m(1+m)},\frac{b}{c}=-\frac{m^2}{(1+m)},\frac{a}{c}=\frac{m}{(1+m{)}^2}.$$ Suppose $m=v/u$, where $v$, $u$ are integers and $\gcd (v,u)=1$. We get $$\frac{a}{c}=\frac{u{v}^2}{v(v+u{)}^2}.$$ This gives $$\frac{a}{u{v}^2}=\frac{c}{v(v+u{)}^2}.$$ Similarly, the expression for $a/b$ gives $$\frac{a}{v^2}=-\frac{b}{u(u+v)}.$$ We thus obtain $$\frac{a}{u{v}^2}=-\frac{b}{u^2(u+v)}=\frac{c}{v(v+u{)}^2}.$$ Let each of these ratios be $p/q$ for some integers $p$, $q$. We obtain $$qa=pu{v}^2,qb=-p{u}^2(u+v),qc=pv(u+v{)}^2.$$ This shows that $\gcd (qb,qc)=\pm p(u+v)$. Thus $$\pm q(b,c)=\pm p(u+v).$$ But $qa=pu{v}^2$. It follows that $\pm q\gcd (a,b,c)=p$. Hence $$\frac{p}{q}=\pm \gcd (a,b,c)=k,$$ say. Finally, we have $$a=ku{v}^2,b=-k{u}^2(u+v),c=kv(u+v{)}^2.$$ It is easy to check that $\{a,b,c\}$ satisfies the given relation.