Dirichlet's theorem

For any two coprime positive integers \(a\) and \(d\), there are infinitely many primes of the form \(a + nd\), where \(n\) is a non-negative integer.

**Argument**: There are infinitely many prime of the form \(n^2 + 1\).

**Proof**:

Let \(n > 0\). For \(n = 1, n^2 + 1 = 2\) which is prime. Otherwise, if \(n^2 + 1 \neq 2\) then \(n^2 + 1 = (n)(n) + 1\) and \(\gcd(n,1) = 1\). Thus, by Dirichlet's theorem there are infinitely many prime of the form \((n)(n) + 1\).

## Comments

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TopNewestWait this is still an unsolved problem right? The proof may seem correct because \(n\) is used in two different contexts in the expression \((n)(n)+1\), one \(n\) is given/fixed while the other \(n\) not. Dirichlet can only tell us that there's infinitely many \(m\) such that \(nm+1\) is a prime, but it doesn't guarentee that \(m=n\). – Xuming Liang · 11 months, 4 weeks ago

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Put n=8 Then from n^2+1 We get 8^2 +1 = 64+1 =65 Which is not prime – Aaryan Saha · 11 months, 2 weeks ago

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– Tapas Mazumdar · 1 month ago

There are infinitely many primes of the form \(n^2+1\) doesn't mean that all the numbers of the form \(n^2+1\) are primes. For example, there are infinitely many positive integers which can be termed as prime numbers and so there are infinitely many primes. Does this mean that all positive integers are prime?Log in to reply