Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$. 
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$

Let $P_{n} = a^{2^n} + b^{2^n} + c^{2^n}$, we have that $$4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$$
$$ = 4(abc)^{2^m} \cdot \left(a^{2^m} + b^{2^m} + c^{2^m}\right) + \left(a^{2^{m + 1}} + b^{2^{m + 1}} + c^{2^{m + 1}}\right)^2 - \left(a^{2^{m + 2}} + b^{2^{m + 2}} + c^{2^{m + 2}}\right)$$
$$ = 4 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right] + 2 \cdot \left[(ab)^{2^{m + 1}} + (bc)^{2^{m + 1}} + (ca)^{2^{m + 1}} \right]$$
$$ = 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$$
So now we need to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$
$\iff \dfrac{1}{2}(a^4 + b^4 + c^4) \mid 4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$. But I don't know about whether it is sufficient to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid P_{n}$ for $\forall n \ge 2$.
And why does $(a, b, c)$ have to be a Pythagorean triple?

HINT.-Given that $a, b, c$ are Pythagorean triples, we can deduce that the problem is equivalent to demonstrating that for every pair of integers $a, b$ the quotient
$$\frac{(a^{2^m}+b^{2^m})(a^2+b^2)^{2^{m-1}}+a^{2^m}b^{2^m}}{(a^2+b^2)^2-a^2b^2}$$ is always an integer for $m\ge2$ (which can be verified numerically).
Trying to apply induction we have for the minimum value $m=2$ the verification (where the numerator is exactly the square of the denominator)
$$\frac{(a^{4}+b^{4})(a^2+b^2)^{2}+a^{4}b^{4}}{(a^2+b^2)^2-a^2b^2}=(a^2+b^2)^2-a^2b^2$$
I cannot for now (lack of time) complete the inductive demonstration so I invite someone who wishes to do so.
