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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.

Given $n$ and $k,$ if there are general formulas for when the coprime condition on $a$ is lifted, which Stangl also solves in the quadratic case, that would be interesting too, but perhaps that is the subject matter for another question. A variant of CRT again boils this down to the prime power case.

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    $\begingroup$ The structure of $(\mathbb{Z}/2^n\mathbb{Z})^*$ is well understood: it is isomorphic to $C_2\times C_{2^{n-2}}$ (when $n\geq 2$), For $k$ odd, the $k$th power map is a bijection so all $2^{n-1}$ possible values of $a$ modulo $2^n$ are $k$th power residues. So you only need to worry about $k=2^j$. $\endgroup$ Aug 20 '20 at 2:20
  • $\begingroup$ @ArturoMagidin that's a neat observation. Any idea what happens in the case that the exponent is a power of $2$? $\endgroup$
    – Favst
    Aug 20 '20 at 2:47
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    $\begingroup$ Clealry, if $j\geq n-1$, then only $1$ is a $2^j$th power modulo $2^n$. For $1\leq j\lt n-1$, the subgroup the image is generated by $5^{2^j}$, and has order $2^{n-1-j}$, so there are exactly $2^{n-1-j}$ residues that are $2^j$th power residues. (Every unit modulo $2^n$ is of the form $\pm 5^k$ for some $k$). $\endgroup$ Aug 20 '20 at 3:43
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    $\begingroup$ Yes; these are all well-known facts. $\endgroup$ Aug 20 '20 at 14:35
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    $\begingroup$ @ArturoMagidin I checked out the revised formula with the powermod(a,b,c) function on WolframAlpha and it works out. In any case, I appreciate the key ideas that you mentioned which allowed me to write up a proof. $\endgroup$
    – Favst
    Aug 21 '20 at 15:37
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Here is the full proof based on ArturoMagidin's comments on the original post (this proof appears in a book that I am writing). Let $k\ge 2$ and $n\ge 3$ be fixed integers. As I wrote in the proof here, the invertible elements modulo $2^n$ are given by the $\varphi(2^n) = 2^{n-1}$ distinct elements of $$S=\left\{\pm 5^k : k\in [2^{n-2}]\right\},$$ where $[t]$ denotes $\{1,2,\ldots, t\}$. Letting $\nu_2 (k)=j,$ we will show that $$ |S_k (2^n)| = \begin{cases} 2^{n-j-2} &\text{ if } 0\le j\le n-3\\ 1 &\text{ if } j\ge n-2 \end{cases}, $$ where $S_k (n)$ denotes the set of reduced $k^{\text{th}}$ power residue classes modulo $n$, meaning those residue classes that are coprime to $n$.

We will use the fact that, for $k\ge 2$, the $k^{\text{th}}$ power map modulo an integer $n\ge 2$ is bijective on the set of reduced residue classes modulo $n$ if and only if $(k,\varphi(n))=1.$ Let $k=2^j m,$ where $m$ is odd. If $j=0,$ then the exponent $k$ is odd and so coprime to $2^n.$ By the aforementioned fact, all integers coprime to $2^n$ are in $S_k (2^n),$ so $|S_k (n)| = 2^{n-2}.$ Now we can assume that $j\ge 1.$ We can think of the $k^{\text{th}}$ power map as taking a power of $m$ of each element of $S$ followed by taking a power of $2^j$ of each element of $S.$ Again by the aforementioned fact, the first map is irrelevant because it is a bijection on $S.$ What really matters is the application of the second map to $$S=\left\{\pm 5^k : k\in [2^{n-2}]\right\}.$$ Since $j\ge 1,$ we can remove any negative signs and we get the set $$T=\left\{ (5^k)^{2^j} : k\in [2^{n-2}]\right\}.$$ Not every $(5^k)^{2^j}$ is necessarily distinct modulo $2^n$ so we have to determine to amount of duplication, or rather the number of distinct elements defined as that will be the cardinality of $S_k (2^n)$. We rewrite each element as $(5^{2^j})^k$ for $1\le k\le 2^{n-2}.$ By the standard proof of the fact that there is no primitive root modulo powers of $2$ (which proves that odd integers have order less than or equal to $2^{n-2}$ modulo $2^n$), we find that $$(5^{2^j})^{2^{n-2}} \equiv (5^{2^{n-2}})^{2^j}\equiv 1^{2^j} \equiv 1\pmod{2^n},$$ and higher powers of $5^{2^j}$ cycle through lower powers. So all elements generated by the powers of $5^{2^j}$ are in $T,$ in addition to every element of $T$ being a power of $5^{2^j}.$ Thus, the number of distinct elements of $T$ modulo $2^n,$ and therefore the cardinality of $S_k (2^n),$ is $$|S_k (2^n)| =\text{ord}_{2^n}(5^{2^j})=\frac{\text{ord}_{2^n}(5)}{(2^j, \text{ord}_{2^n}(5))}=\frac{2^{n-2}}{(2^j, 2^{n-2})}.$$ For $j\ge n-2$ this is equal to $\frac{2^{n-2}}{2^{n-2}} = 1,$ and for $j\le n-3$ this is equal to $\frac{2^{n-2}}{2^j}=2^{n-j-2}.$

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