It's unknown:

1.- if there are an infinite number of primes of the form \(n^2+1\)

2.- if a prime can always be found between \(n^2\) and \((n+1)^2\) (Hardy and Wright 1979, p. 415; Ribenboim 1996, pp. 397-398). These are two of Landau's problems..

This note is written to address these two problems using computers as a tool (creating programs to evaluate, for example, if 2 is true between \(n =1\) to \(n=10^8\), for finding primes of the form \(n^2 +1\) between \(n =1\) to \(n=10^8\)), and with the help of Brilliant Comunity. We are interested in any information or guideline related to these two problem.

For instance, approaching the \(1^{rst}\) \(2^{nd}\) conjeture, I can say : This is just a very, very humble beginning:

1.-

\(\begin{cases}5 = 2^2 + 1 \\ 37 = 6^2 + 1 \\ 101 = 10^2 +1 \\ 197 = 14^2 + 1 \end{cases}\) are prime numbers fullfiling 1, and \(2 , 6, 10, 14\) are in arithmetic progression. Nevertheless, \(325 = 18^2 + 1\) is not a prime number. What happens with \(26^2 + 1\)? What happens with the numbers of the form \((2k)^2 + 1\)?.....

2.-

Between \(1^2\) and \(2^2\) there are 2 primes (2,3).

Between \(2^2\) and \(3^2\) there are 2 primes (5,7).

Between \(3^2\) and \(4^2\) there are 2 primes (11,13).

Between \(4^2\) and \(5^2\) there are 3 primes (17,19,23).

Between \(5^2\) and \(6^2\) there are 2 primes (29,31).

Between \(6^2\) and \(7^2\) there are 4 primes ...

Between \(7^2\) and \(8^2\) there are 3 primes...

Between \(8^2\) and \(9^2\) there are 4 primes...

Between \(9^2\) and \(10^2\) there are 3 primes...

The \(2^{nd}\) conjeture is true for the first 1000 numbers...

I hope this note continue and hopefully we succeed among all.

**P.S**.- I'm not going to answer every questions about this... This is not the proposal of this note.

## Comments

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TopNewest1.- Reductio ad absurdum:

Suppose there are a finite numbers of primes of the form \(n^2 + 1 = 4k^2 +1\). Call \(p\) the greatest prime of the \(4k^2 + 1 \Rightarrow p^2 \) only have 3 factors: \(1, p, p^2 = 16k^4 + 8k^2 + 1\). Of course, \(p^2 + 1\) is not a prime becuase is an even number, and \(p\) is an odd number. Suppose \(p^2 + 1 = 2m \Rightarrow 2m - 1 = p^2 \Rightarrow p^2 - 1 = (p - 1)(p +1) = 2(m - 1)\) \(\Rightarrow m -1 \text{ is an even number }\) \(\Rightarrow m = 2l + 1 \Rightarrow p^2 = 4l + 1 \Rightarrow 4k^4 + 2k^2 = 2k^2(2k^2 + 1) = l \Rightarrow p^2 = 8r + 1\). Now, we have 3 possibilities:

a) \( r \equiv 1 \text{ mod 3 } \Rightarrow p^2\) is divisible by 3 (Contradicition)

b) \( r \equiv 0 \text{ mod 3 } \Rightarrow p^2 = 24q + 1\) with \(r = 3q\). Nevertheless, \( 24 \mid p^2 - 1, \space \forall p > 3\) prime. go here. This implies that \(3 \mid p^2 - 1 \). Conclusion: \(3 \mid p -1 \) or \(3 \mid p +1\). This was obvious...

c) \( r \equiv 2 \text{ mod 3 } \Rightarrow p^2 = 24q + 17 \Rightarrow 24 \mid p^2 - 17\) (contradiction wit th b)). So, we have so far,

\( p = n^2 + 1\) with \(p\) the greatest prime number which can be written of this form, and \(24q = p^2 - 1 = (p - 1)(p + 1) = n^2(n^2 + 2)\). If \(3 | p -1 = n^2\) then \(9 \mid n^2\) because of fundamental theorem of arithmetic and hence \(72 \mid p^2 - 1\) because \( 8 \mid p^2 - 1\) and \(9 \mid p^2 - 1\) with \(\gcd( 8, 9 ) = 1\), or \( 3 \mid n^2 + 2 = 4k^2 + 2 \Rightarrow k \equiv 1 \text{mod 3}\) \(\Rightarrow p = 4k^2 + 1 = 3s + 2\). The diophantine equation \[ 4x - 3y = 1 \] has infintely many solutions of the form \((x = 1 + 3 \lambda, y = 1 + 4 \lambda)\). This implies that \(k^2 = 1 + 3 \lambda \Rightarrow p = 5 + 12 \lambda = 3s + 2\). Let \(p_1 > p \) be a prime number, then there doesn't exist an even natural number \(n_1\) that \(p_1 = n_1^2 +1 = 4a^2 + 1\)

Fermat's theorem on the sum of two squares says:

For odd prime \(p\) \[\exists\ x, y \in \mathbb{Z} \mid p = x^2 + y^2 \] if and only if \[p \equiv 1 \bmod 4\]

There are infinite prime numbers of the form \(4n + 1\). Go here. Let \(p_1 > p\) be a prime of the form \(p_1 = 4n + 1 \Rightarrow p_1 = 4n +1 \neq n_1^2 + 1, \space \forall n_1 \in \mathbb{N} \Rightarrow n_1^2 \neq 4k \Rightarrow k \neq a^2\). This means that an infinite primes greater than \(p\) is of the form \(4k + 1\) then \(k\) is not a perfect square... To be continued – Guillermo Templado · 1 month ago

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