Number of solutions to $a^x+b^y=c^z$ with $\gcd(a,b)>1$

Abstract

We show that there are at most two solutions in positive integers $(x,y,z)$ to the equation $a^x+b^y=c^z$ for positive integers $a$, $b$, and $c$ all greater than one, with just one exceptional case when $\gcd(a,b)=1$, and just one exceptional infinite family of cases when $\gcd(a,b)>1$ (two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are considered the same solution if $\{ a^{x_1}, b^{y_1} \} = \{ a^{x_2}, b^{y_2} \}$). The case in which $\gcd(a,b)=1$ has been handled in a series of successive results by Scott and Styer, Hu and Le, and Miyazaki and Pink, who showed that there are at most two solutions, excepting $(\{a,b\},c) = (\{3,5\},2)$, which gives three solutions. So here we treat the case $\gcd(a,b)>1$, showing that in this case there are at most two solutions, excepting $(a,b,c) = (2^u, 2^v, 2^w)$ with $\gcd(uv,w)=1$, which gives an infinite number of solutions. This generalizes work of Bennett, who proved, for both $\gcd(a,b)=1$ and $\gcd(a,b)>1$, there are at most two solutions $(y,z)$ to the equation $a + b^y = c^z$, and conjectured there are exactly eleven $(a,b,c)$ giving two solutions to this equation (assuming $b$ and $c$ are not perfect powers). For both $\gcd(a,b)=1$ and $\gcd(a,b)>1$, there are an infinite number of $(a,b,c)$ giving two solutions $(x,y,z)$ to the title equation, which are described in detail in this and a cited previous paper. In a further result, in which we no longer say that two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are considered the same solution if $\{ a^{x_1}, b^{y_1} \} = \{ a^{x_2}, b^{y_2} \}$, we list all cases with more than two solutions.