A question concerning higher residues (quadratic and so on)

Question

When $x \equiv a \pmod{n}$ one says that $a$ is the residue of $x$ modulo $n$.

So one can define:

$a$ is a 1-residue modulo $n$ if there is an $x$ with $x \equiv a \pmod{n}$.

Clearly, every $a<n$ is a 1-residue modulo $n$.

When $x^2 \equiv a \pmod{n}$ one says that $a$ is the quadratic residue of $x$ modulo $n$.

So one can define:

$a$ is a 2-residue modulo $n$ if there is an $x$ with $x^2 \equiv a \pmod{n}$.

In general:

$a$ is a k-residue modulo $n$ if there is an $x$ with $x^k \equiv a \pmod{n}$.

And finally

$a$ is a higher residue modulo $n$ when it is a k-residue modulo $n$ for some $k>1$.

In other words:

$a$ is a higher residue modulo $n$ when there is a $k>1$ and an $x$ with $x^k \equiv a \pmod{n}$.

I have learned that

$a$ is a higher residue modulo $n$ iff it is a 1-residue modulo $n$.

– i.e. every $a < n$ is a higher residue.

An argument, why this is so, was given by Darij Grinberg in a comment to another question. It involves Euler's totient function $\varphi$:

You can take $k=\varphi(n)+1$ as $x^{\varphi(n)} \equiv 1 \pmod{n}$ for each $x$ coprime to $n$.

Could anyone please clarify this argument - for me it's too concise.

Answer 1

If $\gcd(a,n)=1$ then $a$ is invertible $\pmod n$, by Bezout's Lemma. Indeed the set of residues coprime with $n$ form a group under multiplication $\pmod n$. By definition the order of that group is $\varphi(n)$. It follows from basic group theory that $\gcd(a,n)=1\implies a^{\varphi(n)}\equiv 1\pmod n$ whence that $a^{\varphi(n)+1}\equiv a\pmod n$

Note, however, that this argument only applies to residue classes prime to $n$. The claim you assert isn't true generally. For example, let $n=4,a=2$. We claim that, for $k>1$, the congruence $x^k\equiv 2\pmod 4$ has no solution. Indeed, we just go case by case. if $x\equiv 0 $ then $x^k\equiv 0$, if $x\equiv 1$ then $x^k\equiv 1$, if $x\equiv 2$ then $x^k\equiv 0$ (since $k>1$), and if $x\equiv 3$ then $x^k\equiv (-1)^k\not \equiv 0$.