Proof that $a^{n}+b^{n}$ is irreducible over $\mathbb Q$

Question

The sum of fourth powers cannot be factored over $\mathbb Q$, since

$ a^4+b^4 = (a^2+\sqrt{2}ab+b^2)(a^2-\sqrt{2}ab+b^2)$

And these quadratic factors does not have any real rational factors.

How to prove that $ a^{n}+b^{n}$ is irreducible over $\mathbb Q$ if $n$ is a power of $2$?

EDIT: This is the problem under considerarion.

Answer 1

Counterexample to the original question: $$(a^4+b^4)(a^8-a^4b^4+b^8)=a^{4\times3}+b^{4\times3}$$ Hint for a possible proof for the revised question: $$b^{2^n}\Phi_{2^{n+1}}\left(\tfrac{a}{b}\right)=a^{2^n}+b^{2^n},$$ where $\Phi_k$ is the $k$-th cyclotomic polynomial, which is irreducible.