+14903 m and n are integers
Hello Guest!
\(m^3 + 24m^2 - 2022 = 3^n\\ .\\ \color{blue}(m,n)\in\{(-17,0),(-15,1),(8,1)\}\\ .\\ ({\color{blue}-17})^3 + 24\cdot ({\color{blue}-17})^2 - 2022 = 3^{\color{blue}0}\\ ({\color{blue}-15})^3 + 24\cdot ({\color{blue}-15})^2 - 2022 = 3^{\color{blue}1}\\ {\color{blue}8}^3 + 24\cdot {\color{blue}8}^2 - 2022 = 3^{\color{blue}1}\\\)
!
+14903 Hi Guest!
\(m^3 + 24m^2 - 2022 = 3^{(n)}\)
I graphed
\(f(m)=m^3 + 24m^2 - 2022 -3^n=0\\ n\in \{0,1,2,\ ...\}\)
The zeros in the graph are approximately
\(m\in \{-17,-15,\ 8\}.\)
By trying with the exponents
\(n\in\{0,1\}\)
I found that the function with the integers
\(m\in \{-17,-15,\ 8\}\)
gives the value zero.
Greatings
!
