36th Hellenic Mathematical Olympiad

Let $y=\frac{p}{q}$, with $p,q\in {\mathbb{N}}^{\ast }$, $(p,q)=1$. Then the equation is written in the form: $$x^q=\frac{y+1}{y}=\frac{\frac{p}{q}+1}{\frac{p}{q}}=\frac{p+q}{p}\Rightarrow x=\sqrt[q]{\frac{(p+q{)}^q}{p^q}}.$$ Therefore the right hand side must be rational. Since $(p+q,p)=1$, it follows that the numerator and the denominator of the fraction must be powers of $p$. Hence $$p+q=a^p\text{and}p=b^p,a,b\in {\mathbb{N}}^{\ast },a>1.$$ If $b\ge 2$, then $b^p\ge 2^p>p$ (absurd), and hence $b=1$, which implies $p=1$ and $q=a-1$. Hence the solutions of the given equation are given parametrically in the form: $(x,y)=\left(a^{a-1},\frac{1}{a-1}\right),a\in {\mathbb{N}}^{\ast },a>1$.