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| Main Authors: | , |
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| Format: | Preprint |
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2024
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| Online Access: | https://arxiv.org/abs/2401.04197 |
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| _version_ | 1866909439606390784 |
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| author | Scott, Reese Styer, Robert |
| author_facet | Scott, Reese Styer, Robert |
| contents | 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. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2401_04197 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | Number of solutions to $a^x+b^y=c^z$ with $\gcd(a,b)>1$ Scott, Reese Styer, Robert Number Theory 11D61 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. |
| title | Number of solutions to $a^x+b^y=c^z$ with $\gcd(a,b)>1$ |
| topic | Number Theory 11D61 |
| url | https://arxiv.org/abs/2401.04197 |