58. National mathematical olympiad Final round

Assume that $a$ and $b$ are not co-prime and let the prime $p$ be their common divisor. We denote by $v_p(n)$ the exact degree of $p$ which divides $n$. Note that $(n,\frac{n^{\ell }-1}{n-1})=1$ for every positive integer $n>1$ and $p$ does not divide $n-1$ if it divides $n$.

Let $(x_1,y_1)$ and $(x_2,y_2)$ be two distinct solutions of the given equation and $x_1>x_2$. Then $b{a}^{x_1}-a{b}^{y_1}+a-b=a^{x_1}-b^{y_1}$ which easily implies that $v_p(a)=v_p(b)$. Subtracting the equalities $\frac{a^{x_1}-1}{a-1}=\frac{b^{y_1}-1}{b-1}$ and $\frac{a^{x_2}-1}{a-1}=\frac{b^{y_2}-1}{b-1}$ we obtain $a^{x_2}\frac{a^{x_1-x_2}-1}{a-1}=b^{y_2}\frac{b^{y_1-y_2}-1}{b-1}$, whence $x_2{v}_p(a)=y_2{v}_p(b)$ and therefore $x_2=y_2$. Now $\frac{a^{x_2}-1}{a-1}=\frac{b^{x_2}-1}{b-1}$ obviously implies that $a=b$, which is a contradiction.