# Prove 2 great theorems (or give a counterexample), using computer as a tool

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. 3 years ago

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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

- 3 years ago