Problem 1399

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.