(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

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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