1

Let $m$ be a square-free number. I want to prove that, given some $b \in \mathbb{Z}^+$, where $\gcd(b,m)>1$, that

$\qquad b^{\,c\,\phi(m) + 1} \equiv b \pmod m,\ $ for some $c \in \mathbb{Z}^+$.

I've verified this with a bunch of different values of $m$ and $b$, but I'm really stuck on how to prove it.

MattAllegro
  • 3,316
user879559
  • 67

1 Answers1

0

Let $b=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, where $p_1, p_2, \ldots p_k$ are prime divisors of b. We'll prove there exist $c$ (actually every natural number works), such that:

$$p_1^{a_1(c\phi(m)+1)} \equiv p_1^{a_1} \pmod {m}$$ $$p_2^{a_2(c\phi(m)+1)} \equiv p_2^{a_2} \pmod {m}$$ $$\cdots$$ $$p_k^{a_k(c\phi(m)+1)} \equiv p_k^{a_k} \pmod {m}$$

Get the first relation and divide with $p_1^{a_1}$ since $m$ is square integer, one is the highest power $p_1$ that devides it. Now we get:

$$p_1^{a_1c\phi(m)} \equiv 1 \pmod n$$

where $n=\frac m{gcd(m,p_1)}$, obviously $(n,p_1) = 1$ since m is squarefree integer. Now use the fact that the Euler Totient Function is multiplicative for relatively prime numbers and that $p_1^{\phi(n)} \equiv 1 \pmod n$:

$$p_1^{a_1c\phi(m)} \equiv p_1^{a_1c\phi(n) \phi(p_1)} \equiv (p_1^{\phi(n)})^{a_1c\phi(p_1)} \equiv 1^{a_1c\phi(p_1)} \equiv 1 \pmod n$$

Simularly we prove for each of the congruence relations.

Now just multiply all the congruence relations and we have:

$$(p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k})^{c\phi(m) + 1} \equiv p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \pmod m$$ $$b^{c\phi(m) + 1} \equiv b \pmod m$$

Hence the proof.

You can directly prove this by using a nice lemma that another MSE user has proved: here. Subsitute $e=c\phi(m) + 1$ in his answer and use that $p-1 = \phi(p)$ and the fact that $m$ is squarefree.

Community
  • 1
Stefan4024
  • 35,843
  • One quick question, what if n is not an integer? Let's say that m = 3511 = 165, and b = 121. n would be 165/121, so how would calculate modulus n. – user879559 Mar 08 '15 at 16:34
  • @user879559 Then you get $n=\frac {165}{11}$. I guess you know that when dividing in congruence relation if the number we divide with doesn't divide the modulo, the modulo doesn't change. So to become clearer divide twice by $11$, after the first dicision you'll have:

    $${11}^{2c\phi(165) + 1} \equiv 11 \pmod {15}$$

    After the second one you'll edn up with:

    $${11}^{2c\phi(165)} \equiv 1 \pmod {15}$$

     – Stefan4024 Mar 08 '15 at 16:38
  • @user879559 You may want to see the new better answer. Also I left you a link to another thing I think may interest you. – Stefan4024 Mar 08 '15 at 17:01
  • @user879559 Well could you tell me why we need to sho what e should be square free? As I've said set $e=c\phi(m) + 1$. For $m=105$, $\phi(m)=48$. Then for $c=6$, we have $e-1=288$. Then obviously $p-1 \mid e-1$ for each prime divisor of $m$. Hence the proof using the lemma – Stefan4024 Mar 08 '15 at 18:09