## (X^4 + Y^4 + Z^4)^2 = 2(X^8 + Y^8 + Z^8)

$(X^4 + Y^4 + Z^4)^2 \; = \; 2 \,(X^8 + Y^8 + Z^8)$

Let’s ignore the trivial solution,   X = Y = Z = 0

Solutions:

$X \; = \; n \,(m^2 - p^2)$
$Y \; = \; 2 \, n \, m \, p$
$Z \; = \; n \,(m^2 + p^2)$

$( \,(n \,(m^2 - p^2))^4 \; + \; (2 \, n \, m \, p)^4 \; + \; (n \,(m^2 + p^2))^4 \,)^2$

$= \; 2 \,( \,(n \,(m^2 - p^2))^8 \; + \; (2 \, n \, m \, p)^8 \; + \; (n \,(m^2 + p^2))^8 \,)$

$= \; 4 \, n^8 \; (m^8 + 14 \, m^4 \, p^4 + p^8)^2$

Note that   (3, 4, 5)   is the least solution