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Prove that for any pair of relatively whole numbers $x$ and $n$, there is some whole number $t$ for which:

$$1+x+x^2...+x^t$$

is a multiple of n An example: $x=168$ and $n=25$, we can find:

$$1+x+x^2+x^3=4,770,025$$

I tried to show that if no $t$ exists, then the series will go on to infinity and converge to the familiar $\frac{1}{1-x}$. I also know for some $r$ and $s$, $ns+xr=1$ (Bezout's identity). I just cant seem to put these two ideas together to prove that there will always be a $t$.

Bill Dubuque
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Randino Andantino mozartino
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2 Answers2

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I don't see any way offhand to use Bézout's identity to prove what you want. Instead, define

$$f(t) = \sum_{i=0}^{t}x^{i} \tag{1}\label{eq1A}$$

Since there are only the $n$ possible remainders of between $0$ and $n - 1$ when any integer is divided by $n$, but more than $n$ values of $f(t)$ among the natural values of $t$ (actually, an infinite number of values), by the Pigeonhole principle, there must be at least $2$ natural numbers $t_1$ and $t_2$, with $t_2 \gt t_1$, such that $f(t_2)$ and $f(t_1)$ have the same remainder when divided by $n$, i.e.,

$$f(t_2) \equiv f(t_1) \pmod n \implies f(t_2) - f(t_1) \equiv 0 \pmod n \tag{2}\label{eq2A}$$

The left side of \eqref{eq2A} is

$$\begin{equation}\begin{aligned} f(t_2) - f(t_1) & = \sum_{i=0}^{t_2}x^{i} - \sum_{i=0}^{t_1}x^{i} \\ & = (1 + x + \ldots + x^{t_1} + \ldots + x^{t_2}) - (1 + x + \ldots + x^{t_1}) \\ & = x^{t_1 + 1} + \ldots + x^{t_2} \\ & = x^{t_1 + 1}(1 + x + \ldots x^{t_2 - t_1 - 1}) \\ & = x^{t_1+1}\sum_{i=0}^{t_2 - t_1 - 1}x^{i} \\ & = x^{t_1+1}f(t_2 - t_1 - 1) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Since $x$ and $n$ are relatively prime, this means that

$$f(t_2 - t_1 - 1) \equiv 0 \pmod n \tag{4}\label{eq4A}$$

giving that $t = t_2 - t_1 - 1$ is your solution.

John Omielan
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  • Can you elaborate ? I don't get this – Randino Andantino mozartino Feb 17 '20 at 08:07
  • @RandinoAndantinomozartino I'm not quite sure what part you don't understand. Nonetheless, I added a detail about how the Pigeon hole principle can be used to show there are at least $2$ with values with the same remainder. Also, I expanded the summations in my ($2$) to show how the lower powers of $x$ cancel, leaving only the higher ones. When the common powers of $x$ are factored out, I get a result which is of the same format at my ($1$), which is your summation. I hope this helps to explain things for you but, if not, please let me know specifically what you don't understand. – John Omielan Feb 17 '20 at 08:15
  • @RandinoAndantinomozartino I just added some details, including inserting a new equation, so my equation ($2$) referenced in my comment above is now ($3$). – John Omielan Feb 17 '20 at 08:21
  • So in the last part because you factor out the x^t+1 , the f ( ...) is divisible by n right? And that means the summation of (t2-t1-1) is a multiple of n ? – Randino Andantino mozartino Feb 17 '20 at 08:25
  • @RandinoAndantinomozartino You are correct. Since $x$ and $n$ and relatively prime, when you have that $x^{t_1 + 1}f(t_2 - t_1 - 1) \equiv 0 \pmod n$, this means that $f(t_2 - t_1 - 1) \equiv 0 \pmod n$, i.e., as you stated, "the summation of (t_2 - t_1 - 1) is a multiple of $n$". – John Omielan Feb 17 '20 at 08:27
  • Can you gmail me ? That way I can ask more questions without posting comments . [email protected] – Randino Andantino mozartino Feb 17 '20 at 08:28
  • @RandinoAndantinomozartino Would you prefer instead switching to chat as we can then discuss things in real time rather than asynchronously as in e-mail? – John Omielan Feb 17 '20 at 08:29
  • Why is f(t1)-f(t2) divisible by n ? – Randino Andantino mozartino Feb 17 '20 at 08:30
  • Great word ! " asynchronously" ..yes let's do that – Randino Andantino mozartino Feb 17 '20 at 08:33
  • Let us continue this discussion in chat. – Randino Andantino mozartino Feb 17 '20 at 08:33
  • Do you understand the Pigeon hold principle? There are only $n$ different remainders between $0$ and $n-1$ inclusive. Thus, if you consider, for example, $t$ going from $0$ to $n$, this gives $n + 1$ remainders between $0$ and $n-1$, inclusive, when divided $f(t)$ is divided by $n$. These cannot all be different, so there are at least $2$ which are the same. I'm letting $t_1$ be the smaller one of these $2$ and $t_2$ being the larger one. Since they have the same remainder when divided by $n$, their difference must be a multiple of $n$, i.e., $f(t_2) - f(t_1)$ is divisible by $n$. – John Omielan Feb 17 '20 at 08:33
  • @Randino If you already know about the order of elements (or Euler's $\phi$ theorem) then you don't need to reinvent the wheel (cycle!) by essentially repeating its pigeonhole proof as in the above answer. Instead, we can simply transform your problem into a problem about the order of $,x,$ as I explain in my answer. – Bill Dubuque Feb 17 '20 at 18:31
  • I'm trying to ' get ' the part about remainders and pigeonhole with the t .. I studied number theory and euler phi function ..is that what this is ? – Randino Andantino mozartino Feb 17 '20 at 19:28
  • @RandinoAndantinomozartino I didn't use the Euler Phi (usually called Totient) function in my answer since I wasn't sure if you had studied it yet. However, you can use it instead, as the answer shows. As for my answer (i.e., using the Pigeonhole principle and remainders) compared to the Totient function, it's not quite the same thing, but some proofs of the function do use similar concepts. – John Omielan Feb 17 '20 at 19:37
  • Why are there more then n values of f(t)? – Randino Andantino mozartino Feb 17 '20 at 19:37
  • @RandinoAndantinomozartino For each $t \ge 0$, there's one value of $f(t)$. Since there's no upper bound on the allowed values of $t$, as it's all non-negative integers, there's always more than $n$ values of $f(t)$ available regardless of how large $n$ is. – John Omielan Feb 17 '20 at 19:39
  • @RandinoAndantinomozartino Perhaps a bit more concrete example will help. Consider the values of $f(t)$ for $0 \le t \le n$, i.e., $n + 1$ values of remainders, say $r$, i.e., with each $0 \le r \le n-1$. If all of the $n + 1$ remainders $r$ were different from each other, there would need to be at least $n + 1$ such different values from among $n$ available remainders. Is that possible? – John Omielan Feb 17 '20 at 19:49
  • Ah I was just going to say before you posted your 2nd to last comment that by taking t ' past ' n , there will be a repeated remainder ..right ? I will clean up these comments after I write them down . You really have taught me something :) – Randino Andantino mozartino Feb 17 '20 at 19:56
  • And above you are counting 0 as one of the " n+1 " remainders right ? – Randino Andantino mozartino Feb 17 '20 at 19:57
  • @RandinoAndantinomozartino For your earlier comment about taking $t$ past $n$ causing a repeated remainder, you are correct. With your comment above, $0$ is one of the possibilities for those "$n + 1$" remainders. Of course, if any are $0$, then you're done as you could use that value. Regardless, though, there will always be at least $2$ which are the same (including both possibly being $0$), which you can subtract to then get a remainder of $0$, with this being the basis for my answer. – John Omielan Feb 17 '20 at 20:02
  • @Randino I added a link in my answer to a Remark giving the simple pigeonhole proof in "order" language. The above pigeonhole proof is essential that proof divided by $,t - 1,,$ since $, f(t) = (x^{\large t+1}-1)/(t-1).\ $ The order viewpoint has the benefit of highlighting the innate group theory. – Bill Dubuque Feb 17 '20 at 21:18
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$$\begin{align} &n\mid \overbrace{(x^{\large t+1}\!-1)/(\color{#0a0}{x\!-\!1})}^{\textstyle x^{\large t}+\cdots +x+1}\\[.3em] \smash{\overset{\!\!\large \times\ (\color{#0a0}{x-1})}\iff}\,\ \ &n(\color{#0a0}{x\!-\!1})\mid x^{\large t+1}-1\\[.4em] \iff\ \ & x^{\color{#c00}{\large t+1}}\equiv 1\!\!\!\pmod{\!m},\,\ m = n(x\!-\!1)\\ \end{align}\ \ \ $$

$x\,$ is coprime to $\,n,x\!-\!1\,$ so such $\,t\!+\!1\,$ exists (by Euler $\phi$ or pigeonhole), e.g. $\,t = {\rm ord}_{m}(x)-1$

E.g. in your example we have that: $\ 168^{\large\color{#c00} 4}\equiv 1\pmod{25\cdot 167},\ $ so $\,\color{#c00}{t\!+\!1=4}\!\iff\! t=3\,$ works.

Bill Dubuque
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