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 · 6 months, 1 week 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 · 6 months ago

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