- #1

tuttlerice

- 28

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Greetings. This is my first post, please be gentle!

I am a music theorist who uses a lot of math in my investigations of music. I am writing a paper about transposition and multiplication operations by 1, 5, 7 and 11 which have very interesting properties in music. The reason, of course, is because those four numbers are coprime to the modulus of 12.

We know because of Dirichlet's theorem that each congruence class entails infinitely many primes, e.g., 12n+1, 12n+5, 12n+7 and 12n+11 each do (although we also know that 12n+1 will entail slightly fewer than the others relative to any given n because, being quadratic, this congruence class supports squares as well as primes, thus taking away from the potential number of instantiations of primes in the congruence class).

My question is this: are there infinitely many primes in the form (n+1)n+1? Dirichlet's theorem says that there are infinitely many primes in the form kn+h where h and k are both integers and coprime. Do we have to know what k actually is? Or do we only have to know that k is an integer and coprime with h? The number 1, of course, is coprime with everything. The way I read the theorem, then, suggests to me that so long as we know that (n+1) and 1 are both integers (in the latter case, we know, and in the former case, we can stipulate), and we know (n+1) and 1 are coprime (they are, whatever n+1 actually is), then we can say with confidence that there are infinitely many primes in the form (n+1)n+1.

However, I am not actually confident of this reading of the theorem, and so I thought I should consult some people who would know better, i.e., you.

Thank you,

Robert Gross, DMA

Shepherd School of Music

Rice University

I am a music theorist who uses a lot of math in my investigations of music. I am writing a paper about transposition and multiplication operations by 1, 5, 7 and 11 which have very interesting properties in music. The reason, of course, is because those four numbers are coprime to the modulus of 12.

We know because of Dirichlet's theorem that each congruence class entails infinitely many primes, e.g., 12n+1, 12n+5, 12n+7 and 12n+11 each do (although we also know that 12n+1 will entail slightly fewer than the others relative to any given n because, being quadratic, this congruence class supports squares as well as primes, thus taking away from the potential number of instantiations of primes in the congruence class).

My question is this: are there infinitely many primes in the form (n+1)n+1? Dirichlet's theorem says that there are infinitely many primes in the form kn+h where h and k are both integers and coprime. Do we have to know what k actually is? Or do we only have to know that k is an integer and coprime with h? The number 1, of course, is coprime with everything. The way I read the theorem, then, suggests to me that so long as we know that (n+1) and 1 are both integers (in the latter case, we know, and in the former case, we can stipulate), and we know (n+1) and 1 are coprime (they are, whatever n+1 actually is), then we can say with confidence that there are infinitely many primes in the form (n+1)n+1.

However, I am not actually confident of this reading of the theorem, and so I thought I should consult some people who would know better, i.e., you.

Thank you,

Robert Gross, DMA

Shepherd School of Music

Rice University

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