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m^3 + 24m^2 - 2022 = 3^n

 

m and n are integers

 May 12, 2022
 #1
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m and n are integers

 

Hello Guest!

 

\(m^3 + 24m^2 - 2022 = 3^n\\ .\\ \color{blue}(m,n)\in\{(-17,0),(-15,1),(8,1)\}\\ .\\ ({\color{blue}-17})^3 + 24\cdot ({\color{blue}-17})^2 - 2022 = 3^{\color{blue}0}\\ ({\color{blue}-15})^3 + 24\cdot ({\color{blue}-15})^2 - 2022 = 3^{\color{blue}1}\\ {\color{blue}8}^3 + 24\cdot {\color{blue}8}^2 - 2022 = 3^{\color{blue}1}\\\)

laugh  !

 May 12, 2022
edited by asinus  May 12, 2022
 #2
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how did you find those values??

Guest May 13, 2022
 #3
avatar+13730 
+1

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

laugh  !

asinus  May 13, 2022

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