## 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     $a \; + \; 2 \, b \; + \; 2 \, c \; = \; 9 \, k$