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Number of solutions to $a^x+b^y=c^z$ with $\gcd(a,b)>1$ (2401.04197v4)

Published 8 Jan 2024 in math.NT

Abstract: We show that there are at most two solutions in positive integers $(x,y,z)$ to the equation $ax+by=cz$ 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 $(a,b,c) = (2u, 2v, 2w)$ 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 $a + by = cz$, 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.

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