$2$ relatively prime numbers and the geometric series 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 $t$.
 A: 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.
A: $$\begin{align}  &n\mid \overbrace{(x^{\large t+1}\!-1)/(\color{#0a0}{x\!-\!1})}^{\textstyle x^{\large t}+\cdots +x+1}\\[.3em]
\smash{\overset{\!\!\large \times\ (\color{#0a0}{x-1})}\iff}\,\ \ &n(\color{#0a0}{x\!-\!1})\mid x^{\large t+1}-1\\[.4em]
\iff\ \ & x^{\color{#c00}{\large t+1}}\equiv 1\!\!\!\pmod{\!m},\,\ m = n(x\!-\!1)\\
\end{align}\ \ \ $$
$x\,$ 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.
