if the repetend length of $b\over a$ being larger or equal than $m-1$ then $m$ is prime?

Question

If : $a,b\in\Bbb{N}$

and we have :

$$a=2^\alpha. 5^\beta.m$$

$$\forall m : (m,10)=1$$

Now how we can show if the period (repetend length) of $b\over a$ being larger or equal than $m-1$ then $m$ is prime?

Example : $a=70$ , $b=3$ $$a=2^1\times5^1\times7$$ $$(7,10)=1$$

$${b\over a}={3\over 70}=0.0\overline{428571}\Rightarrow repetend \, length =6$$ $$6\ge m-1= 7-1 $$ $$\Rightarrow 7 \,is \, \underline{ prime} $$

I'm trying to use Euler's totient function for m and proving that we can divide it by repetend length of $b\over a$ and use this for main question, but i couldn't prove it too.

Thanks in advance.

Answer 1

First we note that factors of $2$ or $5$ in the denominator don't influence the period length.

If $r,s \in \mathbb{N}\setminus\{0\}$ are coprime with $r < s$, and the decimal expansion of $\frac{r}{s}$ has a preperiod of length $\gamma$ and a period of length $\ell$, we have

\begin{align} \frac{r}{s} &= \sum_{k = 1}^{\gamma} \frac{c_k}{10^k} + \sum_{m = 0}^{\infty}\sum_{r = 1}^{\ell} \frac{d_r}{10^{\gamma + r + m\ell}} \\ &= 10^{-\gamma}\underbrace{\sum_{k = 1}^{\gamma} 10^{\gamma-k}c_k}_C + 10^{-\gamma}\underbrace{\sum_{r = 1}^{\ell} 10^{\ell-r}d_r}_D \sum_{m = 0}^{\infty} \frac{1}{10^{(m+1)\ell}} \\ &= \frac{C}{10^{\gamma}} + \frac{D}{10^{\gamma}\bigl(10^{\ell}-1\bigr)} \\ &= \frac{\bigl(10^{\ell}-1\bigr)C + D}{10^{\gamma}\bigl(10^{\ell}-1\bigr)}, \end{align}

so $s \mid 10^{\gamma}\bigl(10^{\ell}-1\bigr)$. Writing $s = 2^{\alpha}\cdot 5^{\beta}\cdot m$ with $\gcd(10,m) = 1$, it follows that $\max \{\alpha,\beta\} \leqslant \gamma$, and $m \mid 10^{\ell}-1$.

Conversely, if $s = 2^{\alpha}\cdot 5^{\beta}\cdot m$ with $\gcd(10,m) = 1$ and $m > 1$, let $\gamma = \max \{\alpha,\beta\}$ and $\ell$ the multiplicative order of $10$ modulo $m$, i.e. $\ell = \min \{ k \in \mathbb{N}\setminus \{0\} : 10^k \equiv 1 \pmod{m}\}$. Then for $u = \frac{10^{\ell}-1}{m}$ we have

$$\frac{r}{s} = \frac{2^{\gamma - \alpha}\cdot 5^{\gamma - \beta}\cdot u\cdot r}{10^{\gamma}\bigl(10^{\ell}-1\bigr)}$$

and with $2^{\gamma - \alpha}\cdot 5^{\gamma - \beta}\cdot u\cdot r = \bigl(10^{\ell}-1\bigr)\cdot C + D$, where $0 < D < 10^{\ell}-1$, we get a decimal expansion with preperiod of length (at most⁽¹⁾) $\gamma$ and period of length $\ell$ reversing the computation above.

So we see that the period length of the decimal expansion is independent of the exponents of $2$ and $5$ in the prime factorisation of the denominator. As long as the numerator and the denominator are coprime, the period length is also independent of the numerator. We summarise:

If $a,b \in \mathbb{N}\setminus \{0\}$ are coprime with $a = 2^{\alpha}\cdot 5^{\beta}\cdot m$ with $\gcd(10,m) = 1$ and $m > 1$, then the (minimal) period length of the decimal expansion of $\frac{b}{a}$ is $$\ell := \min \{ k \in \mathbb{N}\setminus \{0\} : 10^k \equiv 1 \pmod{m}\}.$$

We have $\ell \mid \varphi(m)$ since the order of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ is $\varphi(m)$. In particular, $\ell \leqslant \varphi(m)$. Since $\varphi(m) \leqslant m-1$ for $m > 1$, the period length can never be greater than $m-1$. Now if $\ell = m-1$, then $\varphi(m) = m-1$, and that means $m$ is prime. (For a composite number $n$ with prime factor $p$, we have $\varphi(n) \leqslant n-2$, since $n$ and $p$ are two different numbers in $\{1,2,\dotsc,n\}$ which aren't coprime to $n$.)

Note that the converse doesn't hold, there are lots of primes $p$ for which the period length is smaller than $p-1$.

⁽¹⁾ It is possible that some of the ending digits of $C$ coincide with those of $D$, in which case one could say that the periodic part starts at or before the $\gamma^{\text{th}}$ digit after the decimal point.