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# 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 months, 3 weeks 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 months, 3 weeks ago