Finding the $k^{th}$ root modulo m

Question

We know that method of finding $k^{th}$ root modulo $m$, i.e. if $$x^k\equiv b\pmod m,\tag {$\clubsuit$}$$ with $\gcd (b,m)=1$, and $\gcd(k,\varphi(m))=1$, then $x\equiv b^u\pmod m$ is a solution to $(\clubsuit)$, where $ku-v\varphi(m)=1$. Because $$\begin{array} {}x^k &\equiv \left(b^u\right)^k\pmod m\\ &\equiv b^{uk}\pmod m\\ &\equiv b^{1+v\varphi (m)}\pmod m\\ &\equiv b\cdot b^{v\varphi(m)}\pmod m\\ &\equiv b\cdot \left(b^{\varphi (m)}\right)^v\pmod m\\ &\equiv b\pmod m \end{array}$$

Thus $x\equiv b^u\pmod m$ is a solution to $(\clubsuit)$.

Here we use $\gcd(b,m)=1$, since we used Euler's theorem that $b^{\varphi(m)}\equiv1\pmod m$.

But I am asked to prove that if $m$ is the product of distinct primes, then $x\equiv b^u \pmod m$ is always a solution, even if $\gcd (b,m)\gt1.$

What I did, is say $m=p_1p_2$. Then $\varphi(m)=(p_1-1)(p_2-1)$ $$\begin{array} {}b^{uk}&\equiv b\cdot b^{\varphi (m)}\pmod m\\ &\equiv b\cdot b^{(p_1-1)(p_2-1)}\pmod m \end{array}$$

Now, we just have to compute $b^{(p_1-1)(p_2-1)}\pmod {p_i}$. Here I got stuck, because I really can't use the little theorem for every $p_i$'s, since some $p_i$ can exist in $b$.

Can someone help me?

score 2 · Accepted Answer · answered Dec 27 '16 at 09:20

2

In your case of $m=p_1p_2$. If $\gcd(b,m)>1$, then $\gcd(b,m) \in \{p_1,p_2,m\}$. Without the loss of generality, say $\gcd(b,m)=p_1$. Then \begin{align*} b^{ku} & \equiv b \, . b^{(p_1-1)(p_2-1)}\\ & \equiv 0 \pmod{p_1} & (\because \gcd(b,m)=\gcd(b,p_1)=p_1)\\ & \equiv b \pmod{p_1} \end{align*} Likewise \begin{align*} b^{ku} & \equiv b \, . b^{(p_1-1)(p_2-1)}\\ & \equiv b .1 \pmod{p_2} & (\because \gcd(b,p_2)=1 \text{ so Fermat's little theorem works})\\ & \equiv b \pmod{p_2} \end{align*} Thus $$b^{ku} \equiv b \pmod{p_1p_2}$$

answered Dec 27 '16 at 09:20

Anurag A

41,067

This works much more generally - see my answer. – Bill Dubuque Dec 31 '16 at 22:49
@BillDubuque my intent was not to say this idea works only for two primes. This was just for demo purposes and to answer OP's question. – Anurag A Jan 01 '17 at 03:14
The point of my comment was to inform readers of the relationship between our answers (which may not be immediately obvious), i.e that the above generalizes widely. – Bill Dubuque Jan 01 '17 at 03:18
@BillDubuque That makes perfect sense. – Anurag A Jan 01 '17 at 03:45

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

$b^{\large ku}\!\equiv b\pmod{\!pq}\,$ is case $\,i,j,k=1\,$ of this generalization of the Fermat Euler $\color{blue}{\rm (E)}$ theorem.

${\bf Theorem}\,\ \ n^{\large k+\phi}\equiv n^{\large k}\pmod{\!p^i q^j}\ \ $ if $\,p\ne q\,$ are prime, $ \ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\, $ $\, i,j \le k\ \ \ $

${\bf Proof}\,\ \ p\nmid n\,\Rightarrow\, {\rm mod\ }p^{\large i}\!:\ n^{\large \phi}\!\equiv 1\,\Rightarrow\, n^{\large k + \phi}\equiv n^{\large k},\ $ by $\,\ n^{\large \color{#0a0}\phi} = (n^{\color{#0a0}{\large \phi(p^{\Large i})}})^{\large \color{#0a0}\ell}\overset{\color{blue}{\rm (E)}}\equiv 1^{\large \ell}\equiv 1$

$\qquad\quad\ \ \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^{\large i}\!:\ n^{\large k}\!\equiv 0\,\equiv\, n^{\large k + \phi}\ $ by $\ n^{\large k} = n^{\large k-i} \color{#c00}n^{\large i} = n^{\large k-i} (\color{#c00}{mp})^{\large i}$ and $\,k\ge i$

So $\ p^{\large i}\mid n^{\large k+\phi}\!-n^{\large k}.\,$ By symmetry $\,q^{\large j}$ divides it too, thus so too does their lcm $ = p^{\large i} q^{\large j}\,\ $ QED

Remark $\ $ The above proof immediately extends to an arbitrary number of primes, see this answer. See also Carmichael's Lambda function, a generalization of Euler's phi function.

Finding the $k^{th}$ root modulo m

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