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 n2+1n^2+1

2.- if a prime can always be found between n2n^2 and (n+1)2(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=1n =1 to n=108n=10^8, for finding primes of the form n2+1n^2 +1 between n=1n =1 to n=108n=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 1rst1^{rst} 2nd2^{nd} conjeture, I can say : This is just a very, very humble beginning:

1.-

{5=22+137=62+1101=102+1197=142+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 2,6,10,142 , 6, 10, 14 are in arithmetic progression. Nevertheless, 325=182+1325 = 18^2 + 1 is not a prime number. What happens with 262+126^2 + 1? What happens with the numbers of the form (2k)2+1(2k)^2 + 1?.....

2.-

Between 121^2 and 222^2 there are 2 primes (2,3).

Between 222^2 and 323^2 there are 2 primes (5,7).

Between 323^2 and 424^2 there are 2 primes (11,13).

Between 424^2 and 525^2 there are 3 primes (17,19,23).

Between 525^2 and 626^2 there are 2 primes (29,31).

Between 626^2 and 727^2 there are 4 primes ...

Between 727^2 and 828^2 there are 3 primes...

Between 828^2 and 929^2 there are 4 primes...

Between 929^2 and 10210^2 there are 3 primes...

The 2nd2^{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.

Note by Guillermo Templado
3 years ago

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1.- Reductio ad absurdum:

Suppose there are a finite numbers of primes of the form n2+1=4k2+1n^2 + 1 = 4k^2 +1. Call pp the greatest prime of the 4k2+1p24k^2 + 1 \Rightarrow p^2 only have 3 factors: 1,p,p2=16k4+8k2+11, p, p^2 = 16k^4 + 8k^2 + 1. Of course, p2+1p^2 + 1 is not a prime becuase is an even number, and pp is an odd number. Suppose p2+1=2m2m1=p2p21=(p1)(p+1)=2(m1)p^2 + 1 = 2m \Rightarrow 2m - 1 = p^2 \Rightarrow p^2 - 1 = (p - 1)(p +1) = 2(m - 1) m1 is an even number \Rightarrow m -1 \text{ is an even number } m=2l+1p2=4l+14k4+2k2=2k2(2k2+1)=lp2=8r+1\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) r1 mod 3 p2 r \equiv 1 \text{ mod 3 } \Rightarrow p^2 is divisible by 3 (Contradicition)

b) r0 mod 3 p2=24q+1 r \equiv 0 \text{ mod 3 } \Rightarrow p^2 = 24q + 1 with r=3qr = 3q. Nevertheless, 24p21, p>3 24 \mid p^2 - 1, \space \forall p > 3 prime. go here. This implies that 3p213 \mid p^2 - 1 . Conclusion: 3p13 \mid p -1 or 3p+13 \mid p +1. This was obvious...

c) r2 mod 3 p2=24q+1724p217 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=n2+1 p = n^2 + 1 with pp the greatest prime number which can be written of this form, and 24q=p21=(p1)(p+1)=n2(n2+2)24q = p^2 - 1 = (p - 1)(p + 1) = n^2(n^2 + 2). If 3p1=n23 | p -1 = n^2 then 9n29 \mid n^2 because of fundamental theorem of arithmetic and hence 72p2172 \mid p^2 - 1 because 8p21 8 \mid p^2 - 1 and 9p219 \mid p^2 - 1 with gcd(8,9)=1\gcd( 8, 9 ) = 1, or 3n2+2=4k2+2k1mod 3 3 \mid n^2 + 2 = 4k^2 + 2 \Rightarrow k \equiv 1 \text{mod 3} p=4k2+1=3s+2\Rightarrow p = 4k^2 + 1 = 3s + 2. The diophantine equation 4x3y=1 4x - 3y = 1 has infintely many solutions of the form (x=1+3λ,y=1+4λ)(x = 1 + 3 \lambda, y = 1 + 4 \lambda). This implies that k2=1+3λp=5+12λ=3s+2k^2 = 1 + 3 \lambda \Rightarrow p = 5 + 12 \lambda = 3s + 2. Let p1>pp_1 > p be a prime number, then there doesn't exist an even natural number n1n_1 that p1=n12+1=4a2+1p_1 = n_1^2 +1 = 4a^2 + 1

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

For odd prime pp  x,yZp=x2+y2\exists\ x, y \in \mathbb{Z} \mid p = x^2 + y^2 if and only if p1mod4p \equiv 1 \bmod 4

There are infinite prime numbers of the form 4n+14n + 1. Go here. Let p1>pp_1 > p be a prime of the form p1=4n+1p1=4n+1n12+1, n1Nn124kka2p_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 pp is of the form 4k+14k + 1 then kk is not a perfect square... To be continued

Guillermo Templado - 3 years ago

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