I'm trying to find the number of $n$ x $n$ encrypting matrices modulo 26 for $n$ $\in$ $\Bbb N$.
I really don't know what method to use to approach this... does it involve Hill ciphers?
Number of encrypting $n\times n$ matrices modulo 26
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$\begingroup$
linear-algebra
matrices
cryptography
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1Why are you trying to find this? What does the question relate to? Further context is needed IMO. – 2017-02-09
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0The question is "Count the number of encrypting n-by-n matrices modulo 26, for $n \in \Bbb N$". The context is that it is Cryptography, no other information was given, just count the number of ways. – 2017-02-09
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0The number of $n\times n$ invertible matrices modulo $26$ is the product of the number of invertible matrices modulo $2$ and the number of invertible matrices modulo $13$. Those are easier to count. However, it isn't obvious if you want more than invertible for it to be "encrypting." – 2017-02-09
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0You need more context and information in the question imo. What you are asking is not enough for those who are not totally familiar with the topic. – 2017-02-09
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0While someone may have passed such a problem to you, and you then pass it through to us, originally the notion of "encrypting matrices modulo $26$" had to be defined. Perhaps you are not studying Cryptography from a specific book, but Math.SE Readers depend on you as the one posting the Question to be expert on what the Question is about. – 2017-02-10
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0@JonathanBa You should help people help you. – 2017-02-10
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0See http://jeff.over.bz/papers/undergrad/on-the-keyspace-of-the-hill-cipher.pdf – 2017-02-12
1 Answers
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The number of $n\times n$ invertible matrices modulo a prime $p$ is:
$$(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})=p^{n(n-1)/2}(p-1)(p^2-1)...(p^n-1)$$
The number of invertible matrices modulo $26$ is the product of the numbers for $p=2$ and $p=13$. (This is an application of the Chinese remainder theorem to matrices.)
So that gives you:
$$26^{n(n-1)/2} (13^1-1)(2^1-1)\cdot(13^2-1)(2^2-1)\cdot(13^{n-1}-1)(2^{n-1}-1)$$