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I'm trying to prove a fact about $k$th roots that says that if $(k, \phi(m)) = 1$ and $(b,m) = 1$, then there is a unique $k$th root modulo $m$ (composite $m$).

I'm not sure how to go about it but here's an attempt;

Suppose towards a contradiction that there are two solutions $x_1 = b^r$ and $x_2 = b^s$ to the congruence $x^k \equiv b \bmod m$. Then $$x_1^k \equiv x_2^k \equiv b \bmod m$$

so that $b^{kr} \equiv b^{ks} \equiv b \bmod m$, or equivalently, $b^{kr - 1} \equiv b^{ks - 1} \bmod m$. We can then conclude that $m \mid b^{kr - 1} - b^{ks - 1}$. Now we can factor to obtain $b^{kr - 1} - b^{ks - 1} = b^{kr - 1}(b^{k(s-r)} - 1)$ to conclude that $m \mid b^{kr - 1}(b^{k(s-r)} - 1)$. But since $(b, m) = 1$, we must have $m\mid b^{k(s-r)} - 1$, or equivalently,

$$b^{k(s-r)} \equiv 1 \bmod m$$

which would imply $k(s-r) = \phi(m)v$ for some $v \in \mathbb Z$.

I feel like there could be a way to extract a contradiction from this last step but I can't see it, would somebody be able to check if this is a good way to go?

Edward Evans
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  • Why do you presume that the solutions are powers of $b?\ $ – Bill Dubuque Aug 07 '17 at 13:42
  • @BillDubuque I'm not sure, I think just because the algorithm described in the textbook I'm using takes $x = b^u$ as a solution, with $ku - \phi(m)v = 1$. – Edward Evans Aug 07 '17 at 13:50
  • The textbook is probably using something similar to the methods in the answers below. If your proof is a continuation of the textbook proof then it is essential that you supply the initial parts of the proof. – Bill Dubuque Aug 07 '17 at 13:52
  • @BillDubuque The textbook provides no proof, it's an exercise at the end of the chapter so I'm simply guessing that $x$ is a power of $b$ in the same way that the algorithm described says that $x$ is a power $b$, if that helps. – Edward Evans Aug 07 '17 at 13:54
  • What precisely is the algorithm described in the textbook? – Bill Dubuque Aug 07 '17 at 13:57
  • @BillDubuque Compute $\phi(m)$, find positive integers $u, v$ such that $ku - \phi(m)v = 1$, compute $b^u \bmod m.$ This is a solution to the congruence $x^k \equiv b \bmod m$, since $x^k \equiv b^{uk} \equiv b^{1 + \phi(m)v} \equiv b \bmod m$ by Euler's formula. – Edward Evans Aug 07 '17 at 14:00
  • i.e. $! \bmod{\phi}!:,\ ku\equiv 1,,$ so $,u\equiv 1/k.,$ This is the same idea used in all the answers below. In my answer I tried to make it explict how it amounts to taking $k$'th roots - which is the same as taking a $u$'th power, since the "fraction" $1/k := k^{-1}$ is congruent to the integer $,u\pmod{\phi}\ \ $ – Bill Dubuque Aug 07 '17 at 14:05

3 Answers3

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Here is a simpler approach:

Since $(k, \phi(m)) = 1$, write $uk + v \phi(m)=1$ for $u,v \in \mathbb Z$.

Then, if $(x,m)=1$ we have $x=x^1=x^{uk + v \phi(m)}={(x^u)}^k (x^{\phi(m)})^v \equiv {(x^u)}^k \bmod m$.

This proves that the map $x \mapsto x^k$ is surjective and so is a bijection.

In other words, every $x$ with $(x,m)=1$ has a unique $k$-th root mod $m$.

lhf
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To (uniquely) solve $\ x^{\large k} \equiv b\ $ take a $k$'th root (i.e. raise both sides to power $1/k)$, just as for reals.

Let $\phi = \phi(m).\, $ $\,(k,\phi) = 1\,\Rightarrow\, j := 1/k\pmod{\!\phi}\,$ exists, so raising to power $\,j\,$ yields

$$\begin{align} &\left[x_1^{\large k} \equiv b\equiv x_2^{\large k}\right]^{\large j}\\ \Rightarrow\ &\ \ x_1 \equiv b^{\large j}\!\equiv x_2\end{align}$$

This works because by Euler all exponents can be taken mod $\phi$ (on all $\,a\,$ coprime to $m),\,$ i.e. $\!\bmod m\!:\ \color{#c00}{a^{\large \phi}\equiv 1}\,\Rightarrow\, a^{\large r+q\,\phi}\equiv a^{\large r}(\color{#c00}{a^{\large \phi}})^{\large q}\equiv a^{\large r}\color{#c00}1^{\large q}\equiv a^{\large r},\ $ i.e. $\,a^{\large n}\!\equiv a^{\large n\bmod\phi},\,$ hence $$ (x^{\large k})^{\large j}\! \equiv x^{\large kj}\!\equiv x^{\large kj\bmod\phi}\!\equiv x^{\large k(1/k)\bmod \phi}\!\equiv x^{\large 1}$$ Thus the solution amounts to raising $\,x^{\large k}\,$ to power $\,1/k\,$ (i.e. taking its $k$'th root) to solve for $\,x.$

Note: it is valid to apply Euler $\phi\,$ to powers of $x$ since $\,(x,m) = 1,\,$ by $\,1=(b,m)=(x^k,m)$

See this answer for elaboration and a fully worked example.

Bill Dubuque
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  • Thanks for the answer. I managed alone to get to the statement that $k(s-r) = \phi(m)v$ for some $v \in \Bbb Z$. From this I can deduce that $s-r = \phi(m)u$ for some $u \in \Bbb Z$ so that $b^{s-r} \equiv 1 \bmod m$ and so that $x_1 \equiv x_2 \bmod m$, contradicting the original assumption. Is this in any way logically flawed or does this work to show that the congruence has a unique solution? Not to disregard your answer, I'd just also like to have my own proof verified! :) – Edward Evans Aug 07 '17 at 13:39
  • Ok so in my case I've got $k=2019$ but the answer is not $b^{1/2019} (\text{mod} \ p)$ but rather $b^\alpha \ (\text{mod} \ p)$ according to my original thread. Also, you use $\phi$, which I assume is Eulers totient function? But totient function of what? Why is this true: $,(k,\phi) = 1,\Rightarrow, j := 1/k,$ exists $!\pmod{!\phi}\ \ ?$ I really can't seem to grasp how to apply this to my problem. – Parseval Feb 18 '19 at 21:28
  • @Parseval As above, $\ j = 1/k\bmod \phi = 1/2019\bmod p!-!1,,$ by $,\phi(p) = p!-!1.,$ Recall by Bezout $,\gcd(k,\phi) = 1,\Rightarrow, \exists,j,n!:\ jk+n\phi = 1,\Rightarrow\bmod \phi!:\ jk\equiv 1,\Rightarrow, j\equiv 1/k.,$ – Bill Dubuque Feb 18 '19 at 21:39
  • @BillDubuque - Ok I think I'm starting to get things now. I only need further elaboration on the part where you write "...all exponents can be taken mod $\phi$..." and how you from that conclude that $a^n\equiv a^{n \ \text{mod} \ \phi}$. – Parseval Feb 19 '19 at 13:54
  • @Parseval That's what we prove in the subsequent equations: i.e. if $,a^{\large \phi}\equiv 1\pmod{!m},$ then $,n\equiv r\pmod{!\phi},\Rightarrow$ $,a^{\large n}\equiv a^{\large r}\pmod{!m}.,$ In particular that holds when $,r = (n\bmod \phi).\ $ – Bill Dubuque Feb 19 '19 at 15:13
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There is $l$ with $kl\equiv 1\pmod{\phi(m)}$. Note that for all $a$ coprime to $m$, then $r\equiv s\pmod{\phi(m)}$ implies $a^r\equiv a^s\pmod m$.

Then $$b\equiv a^k\pmod{m}\iff a\equiv b^l\pmod{m}.$$ One way: $b\equiv a^k\pmod{m}\implies b^l\equiv a^{kl}\pmod m$. As $a^{kl}\equiv a\pmod{m}$ (why?) that does it. Other way: similar.

Angina Seng
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