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Let $m, n \in \mathbb{Z}_+$ with $n > 2$, and let $\frac{n^m-1}{n-1}$ be divisible by $17$. Show that either $m$ is even:$ m \equiv 0 \mod 17$ and $n \equiv 1 \mod 17$. Find all possible values of $n$ in the cases when $m$ is even but not divisible by $4$, or divisible by $4$ but not divisible by $8$.

So far, I have done this: First, let's assume that $n \equiv 1 \mod 17$. Also, we know that we can write

$$\frac{n^m-1}{n-1} = \sum_{i = 0}^{m-1} n^i.$$

As $n \equiv 1 \mod 17 \implies n^i \equiv 1 \mod 17$ and so $\sum_{i = 0}^{m-1} n^i \equiv \underbrace{1 + 1 + \cdots + 1} \equiv m$.

Also, we know that $\sum_{i = 0}^{m-1} n^i \equiv 0 \mod 17 \implies m \equiv 0 \mod 17$.

Now, let's assume that $n \not\equiv 1 \mod 17$. Then, $\frac{n^m - 1}{n - 1}$ is divisible by $17$ if $n^m - 1$ is divisible by $17$, i.e $n^m - 1 \equiv 0 \mod 17 \implies n^m \equiv 1 \mod 17$. Fermat's little theorem tells us that $n^{p-1} \equiv 1 \mod p \implies n^{16} \equiv 1 \mod 17$. From here, we have that $m \mid 16$ which tells us that $m$ must be even.

However I am now stuck on the next bit, I'm not sure how to find those values of $n$. Can someone help me please. Thank you.

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up vote 1 down vote accepted

You have made considerable progress, indeed you are near the end. If $n\equiv 1\pmod{17}$, then by what you wrote we have $m$ must be divisible by $17$, and any $m$ divisible by $17$ will do. The condition $m$ divisible by $2$ but not by $4$ is easy to meet. We need $m\equiv{34}\pmod{68}$. Then any $n\equiv 1\pmod{17}$ will do the job.

Now let us look at the case $n\not\equiv 1\pmod{17}$. So $n-1$ and $17$ are relatively prime. So our ratio is divisible by $17$ if and only if $n^m\equiv 1\pmod{17}$. This automatically holds if $16$ divides $m$. But our condition on $m$ is that $m$ is divisible by $2$ but not by $4$, so we are far from $16$ divides $m$.

The order of $n$ modulo $17$ divides $16$, so is a power of $2$. We want the order to be exactly $2$. This is the case if and only if $n\equiv -1\pmod{17}$.

For the second part ($m$ divisible by $4$ but not by $8$), there will be some work left to do. You need to determine which $n$ have order $4$ modulo $17$.

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Think you misread it slightly lol, I need the $n$ values, not the $m$ ones. Unless that was a typo on your part? – Kaish May 24 '13 at 19:40
@Kaish: For $m$ of the right shape, any $n$ congruent to $1$ modulo $17$ works. In the material I added, there is a discussion (with proof) of the $n$ that work if $n\not\equiv 1\pmod{17}$. – André Nicolas May 24 '13 at 19:49

As the first case is perfect, let me start with the second.

As $17$ is prime and hence $\phi(17)=16$

As Primitive Root is a Generator of Reduced Residue System, let's find a primitive root of $17$

$2^4\equiv-1, 2^8\equiv(-1)^2\equiv1\pmod{17} \implies \text{ord}_{17}2=8<\phi(17)$

$3^2\equiv9, 3^4\equiv-4,3^8=(3^4)^2\equiv(-4)^2\equiv-1\pmod {17}$ $\implies \text{ord}_{17}3=16$ i.e., $3$ is a primitive root of $17$

From this, if $d=$ord$_ma,$ ord$_m(a^r)=\frac d{(r,d)}$ where $m,d$ are positive integers and $a,r$ are integers.

So, ord$_{17}(3^n)=\frac{16}{(16,n)}$

Case $1:$ If ord$_{17}(3^n)\equiv2\pmod 4, n=8s$ where $s$ is odd $\implies 3^{8s}=(3^8)^s\equiv(-1)^s\equiv-1\pmod{17}\implies $ord$_{17}(-1)=2$

Case $2:$ If ord$_{17}(3^n)\equiv4\pmod 8, n=4s$ where $s$ is odd $\implies n\equiv4,12\pmod{16}$

Now, $3^4\equiv-4\pmod{17}$ and $3^{12}=3^8\cdot3^4\equiv(-1)(-4)\equiv4\pmod {17}$ $\implies $ord$_{17}(\pm4)=4$

Case $3:$ If ord$_{17}(3^n)\equiv8\pmod {16}, n=2s$ where $s$ is odd $\implies n\equiv2,6,10,14\pmod{16}$

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