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Given an LCG with parameters $a$,$c$,$m$, prove that $$x_{n+k}\equiv a^{k}x_n + \frac{(a^{k} - 1)}{a-1}c \mod m; \ (a \geq 2, k\geq 0)$$ which shows that the $(n+k)\text{th}$ term can be computed directly from the $n\text{th}$ term.

Attempted proof - Suppose we have an LCG $$x_n\equiv a x_{n-1} + c \mod m$$ where $a$ is called the multiplier, $c$ is called the shift or increment, and $m$ is called the modulus of the generator. I am not really sure how to prove the latter above, perhaps I need to multiply $x_n$ by itself $k$ times, and proceed inductively. For example, let's say $x_{n+k}$ above holds, then for $k = 1$ we have $$x_{n} \equiv a x_{n} + c \mod m$$ I am just struggling with proving these types of problems...

As an aside I never taken a number theory course, I just found this exercise in a pdf I found online and wanted to give it a try. Any recommendation are greatly appreciated. Also, a detailed proof of this should help me with others I would like to attempt.

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Try induction in $k$ for example. $k=0$ reduces to $x_n=x_n$ which obviously holds true. Next, assume it holds true for $k$ which means:

$$ x_{n+k} = a^{k}x_n + \frac{a^{k} - 1}{a-1}\,c $$

The following proves that it must hold true for $k+1$ as well, which completes the induction step:

$$ \begin{align} x_{n+k+1} & = a x_{n+k} + c \\ & = a\left(a^{k}x_n + \frac{a^{k} - 1}{a-1}\,c\right) + c \\ & = a^{k+1}x_n + \left(a \cdot \frac{a^{k} - 1}{a-1} + 1\right) c \\ & = a^{k+1}x_n + \frac{a^{k+1} - a + a - 1}{a-1} \,c \\ & = a^{k+1}x_n + \frac{a^{k+1} - 1}{a-1} \,c \end{align} $$

The above proves the relation in whole numbers. It will still hold if you replace $=$ with $\equiv \bmod m$.

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    I am not following completely, I see for $k = 0$ it holds true and the same follows for $k = 1$. Then in your next line and the ones there after I am a bit lost. – 2017-01-27
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    Also when you set $k = 0$ it seems that we have $$x_{n} = x_{n}\mod m$$ what happens to the $m$ for which we have $x_n = x_n$? – 2017-01-27
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    @Scooby Edited to clarify the induction steps. As for the second question, $x_n = x_n \implies x_n = x_n \bmod m$ but $x_n = x_n$ is a stronger statement. – 2017-01-27
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    Ok, I understand. Now when you assume it holds true for $k$, you next try to show that it must hold for $k+1$? – 2017-01-27
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    @Scooby Yes, and the second line uses the expression for $x_{n+k}$ known to be true by the hypothesis of the induction step. – 2017-01-27
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    Ok, I understand now. So it suffies as a complete prove to first show that it holds for $k=0$ then we assume it holds for $k$ and prove it holds for $k+1$ and then we are done? – 2017-01-27
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    @Scooby Yes, that's how [induction](https://en.wikipedia.org/wiki/Mathematical_induction#Description) works. – 2017-01-27
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    Great, thank you very much. – 2017-01-27