We define a $k^{\text{th}}$ power residue modulo $n$ to be an integer $a$ coprime to $n$ such that there exists an integer $x$ satisfying $$x^k\equiv a\pmod{n}.$$ A fundamental question that we can ask is:
Given fixed integers $k\ge 2$ and $n\ge 2,$ how many $k^{\text{th}}$ power residues modulo $n$ exist?
A variant of the Chinese remainder theorem for $k^{\text{th}}$ powers shows that this is the product of the answers when the moduli are the the maximal prime powers in the prime factorization of $n.$ For powers of odd primes $p,$ this is solved by a variant of Euler's criterion (see section 6.5 of Vinogradov's Elements of Number Theory).
But what about when the modulus is a power of $2$? Unfortunately, Vinogradov's proof in the odd prime case uses primitive roots, which does not exist modulo powers of $2$ that are greater than $4=2^2$. In the quadratic case, a paper of Stangl provides a formula on p.287-288. His method uses the difference of squares factorization and other ideas that do not readily extend to higher powers, at least not as far as I can tell.
Please let me know if there is a formula for the number of $k^{\text{th}}$ power residues modulo $2^n$ for fixed integers $k\ge 2$ and $n\ge 3$.
The answer might have to do with the structure of $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ but I'm not sure.