fractional part of (6*sqrt(6) + 14)

 
 

A \; = \; (6 \, \sqrt {6} \; + \; 14)

frac(6 \, \sqrt {6} \; + \; 14) \; \times \; (6 \, \sqrt {6} + 14) \; = \; 20

 

A_2 \; = \; (6 \, \sqrt {6} \; + \; 14)^2

frac(6 \, \sqrt {6} \; + \; 14)^2 \; \times \; (6 \, \sqrt {6} + 14)^2 \; = \; 20^2

 

A_3 \; = \; (6 \, \sqrt {6} \; + \; 14)^3

frac(6 \, \sqrt {6} \; + \; 14)^3 \; \times \; (6 \, \sqrt {6} + 14)^3 \; = \; 20^3

 

A_4 \; = \; (6 \, \sqrt {6} \; + \; 14)^4

frac(6 \, \sqrt {6} \; + \; 14)^4 \; \times \; (6 \, \sqrt {6} \; + \; 14)^4 \; = \; 20^4

 

Prove that if    N \; = \; (6 \, \sqrt {6} \; + \; 14)^{2n + 1},   then

frac(6 \, \sqrt {6} \; + \; 14)^{2n+1} \; \times \; (6 \, \sqrt {6} \; + \; 14)^{2n + 1} \; = \; 20^{2n + 1}

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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