$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ 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!
 A: Bear with me as I ramble a bit ...
The relation in question resembles a way of writing Heron's formula for the area of a triangle with sides $x$, $y$, $z$:
$$16\;|\triangle xyz|^2 = \left(x^2+y^2+z^2\right)^2-2\left(x^4+y^4+z^4\right) \tag{1}$$ Now, some (most?) people think of Heron's formula as more like
$$|\triangle xyz|^2 = s(s-x)(s-y)(s-z) \tag{2}$$
where $s = (x+y+z)/2$ is the semi-perimeter. I personally prefer not to bother introducing an extra value, so I write
$$16\;|\triangle xyz|^2 = (x+y+z)(-x+y+z)(x-y+z)(x+y-z) \tag{3}$$
What's the point of all this? Well, it actually has nothing to do with triangle areas. My point is simply that my familiarity with various forms of Heron formula allows me to immediately realize that expressions that look like $(1)$ factor into expressions that look like $(3)$. (So, I no longer need to go through the trouble of expanding the product, combining terms, and attempting to factor. It's second-nature to me now, and should be to any olympiad contender. :) The same must be true of the relation in question:
$$\begin{align}
0 &= \left(a^4+b^4+c^4\right)^2-2\left(a^8+b^8+c^8\right) \\[4pt]
&= \left(a^2+b^2+c^2\right)\left(-a^2+b^2+c^2\right)\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)
\end{align} \tag{4}$$
Assuming this relation holds, we see that one of the factors must vanish. Obviously, first never does. Whichever of the latter three does corresponds to a Pythagorean relation in $a$, $b$, $c$, and therefore a right triangle. Conversely, a right triangle admits a Pythagorean relation, which causes a $(4)$ to hold. $\square$
A: 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.
