$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ if and only if $a,b,c$ are the lengths of a right angled triangle Here is a problem from the book: Everything connected to Pithagoras
It is well known that in every right angled triangle $ABC$, $a^2+b^2=c^2$. Howevee, there are some more complicated equations as well. Here is one:
Prove that $$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$$ if and only if $a,b,c$ are the lengths of a right angled triangle.
I can’t prove it. I tried to write $a_4=x$ and so on, but only algebra didn’t help for me. Aldo it is interesting because in this problem it doesn’t matter which side is the hypotenuse...
Please help!

Hint:
$$
2(a^4b^4+b^4c^4+c^4a^4)-(a^8+b^8+c^8)\\
=(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).
$$

Edit: I arrived at the above by guessing. My intuition is that: the given identity $2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ is symmetric in its info about $a$, $b$, and $c$, therefore, the conclusion has to be symmetric about $\angle A$, $\angle B$, and $\angle C$. So a product
$$
(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)
$$
came to mind. But it is 2 degrees lower than the degree of the given identity. So  I needed a 2nd degree factor. Again, that factor has to be symmetric in $a$, $b$, and $c$. So $a^2+b^2+c^2$ was the immediate candidate (I would have tried $(a+b+c)^2$ next if $(a^2+b^2+c^2)$ hadn't worked.). Then, it remained for me to expand and simplify
$$
(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)
$$
using Mathematica.
