a^n + b^n + c^n = d^n + e^n + f^n, n =1,2 — (Part 2)

 

 

a \; + \; b \; + \; c \; = \; (k - a) \; + \; (k - b) \; + \; (2 \, k - c)
a^2 \; + \; b^2 \; + \; c^2 \; = \; (k - a)^2 \; + \; (k - b)^2 \; + \; (2 \, k - c)^2

Since,

a \; + \; b \; + \; c \; = \; (k - a) \; + \; (k - b) \; + \; (2 \, k - c)

then,

a \; + \; b \; + \; c \; = \; -a \; - \; b \; - \; c \; + \; 4 \, k
2 \, a \; + \; 2 \, b \; + \; 2 \, c \; - \; 4 \, k \; = \; 0
a \; + \; b \; + \; c \; = \; 2 \, k

Then,

(a^2 + b^2 + c^2) \; - \; ((k - a)^2 \; + \; (k - b)^2 \; + \; (2 \, k - c)^2) \; = \; 0
2 \, a \, k \; + \; 2 \, b \, k \; + \; 4 \, c \, k \; - \; 6 \, k^2 \; = \; 0
k \; = \; (1/3) \; (a + b + 2 \, c)

All that is required is to select   a, \; b, \; c   so that   (a + b + 2 \, c)   is divisible by 3

Note that

a^2 \; + \; b^2 \; + \; c^2 \; = \; (2 \, k - a)^2 \; + \; (4 \, k - b)^2 \; + \; (4 \, k - c)^2

when     a \; + \; 2 \, b \; + \; 2 \, c \; = \; 9 \, k

In summary:

a^2 \; + \; b^2 \; + \; c^2 \; = \; (k - a)^2 \; + \; (k - b)^2 \; + \; (2 \, k - c)^2

when     a \; + \; b \; + \; 2 \, c \; = \; 3 \, k

 

a \; + \; b \; + \; c \; = \; (2 \, k - a) \; + \; (4 \, k - b) \; + \; (4 \, k - c)
a^2 \; + \; b^2 \; + \; c^2 \; = \; (2 \, k - a)^2 \; + \; (4 \, k - b)^2 \; + \; (4 \, k - c)^2

when     a \; + \; 2 \, b \; + \; 2 \, c \; = \; 9 \, k

 

 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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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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