2009/029) prove a+b+c=0 ==> 2a⁴ +2b⁴ +2c⁴ = n²

a+b+c = 0

the a+b = - c
square both sides

(a^2+b^2+2ab) = c^2
so (a^2+b^2) = c^2 -2ab

now again square both sides

a^4+b^4 + 2a^2b^2 = (c^2-2ab)^2
or a^4 + b^4 + 2a^2b^2 = c^4 + 4a^2b^2 - 4abc^2
or a^4+b^4 = c^4 + 2a^2b^2 - 4abc^2
add c^4 on both sides to get

a^4+b^4+c^4 = 2c^4 + 2a^2b^2 - 4abc^2

or 2(a^4+b^4+c^4) = 4c^4+4a^2b^2 - 8abc^2
= 4(c^4-2abc^2+a^2b^2)
= 4(c^2-ab)^2

which is a perfect quare putting n = 2 | c^2-ab|
proved