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This is a statement about Repunits from this paper.

How can I write this as an if/then statement?
Knowledge about Repunits isn't required. The question is basically: what does "although it is necessary for $a$ to be prime in order for $b$ to be prime" mean?

I think it means: if $b$ is prime then $a$ is prime.

Taroccoesbrocco
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Michael Romero Jr
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  • "$P$ is necessary for $Q$" (or equivalently, "$Q$ is sufficient for $P$") is, more formally, $Q\to P$. However, linguitically, I find the twofold use of "for" in your quote confusing – Hagen von Eitzen Oct 14 '18 at 20:08

2 Answers2

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although it is necessary for "$a$ to be prime" in order for "$b$ to be prime".

That is, if $b$ is prime, then we know that $a$ must be prime.

You are right that the right translation is

"If $b$ is prime, then $a$ is prime" which corresponds to "If $R_n$ is prime, then $n$ is prime"

Notice that in this case we say that "$a$ is prime" is a necessary condition to "$b$ is prime". We say that "$b$ is prime" is a sufficient condition for "$a$ is prime".

Siong Thye Goh
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The logically equivalent statement to yours might be:

If $R_n$ is prime then $n$ is prime.

If you are interested in proving such statement, consider the following "proving the contrapositive":

$R_n$ can be written as $\frac{10^n-1}{9}$ given that the fraction $\frac{10^n-1}{9}$ is always an integer. Now, if $n$ is not prime, then $n = ab$ for some prime $a$ and integer $b$. Hence, $R_n = \frac{10^{ab}-1}{9}=\frac{(10^a)^b-1}{9}$. But $(10^a)^b-1 = (10^a - 1) \times \sum_{i=0}^{b-1}(10^a)^{i}$ and $R_n = \sum_{i=0}^{b-1}(10^a)^{i} \times \frac{10^a - 1}{9}$ This implies that $(\sum_{i=0}^{b-1}(10^a)^{i})$ is a divisor of the Repunit $R_n$. Thereby, if $n$ is not prime, then so is $R_n$. Contrapositively, $n$ must be prime if $R_n$ is prime.

You may read this post for more details about repunits.

Maged Saeed
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