Prime Numbers in the Diophantine equation q=(n^2+1)/p and p is Prime

AntonVrba

Investigating the Diophantine equation [tex]q = \frac{n^2+1}{p}}[/tex] where [tex]{p}[/tex] is a prime number, [tex]n,q[/tex] are integers per definition

The prime numbers can be sorted into two groups

Group 1 has no solution and

Group 2 has the solution [tex] n = \{ a\times p - b{ \ },{ \ } a\times p + b \} {\ \ \ }\forall { \ \ }a>=0[/tex]

The table below list results for the first view primes, there is no particular pattern which divides the primes into either group 1 or 2 nor a pattern for the value [tex]b[/tex] and there seem to be an equal number group1 and group2 primes.
[tex]\begin{array}{cc,c,c}
{No.&Group\ 1&Group\ 2&b\\
1&{}&2&1\\
2&3&{}&{}\\
3&{}&5&2\\
4&7&{}&{}\\
5&11&{}&{}\\
6&{}&13&5\\
7&{}&17&4\\
8&19&{}&{}\\
9&23&{}&{}\\
10&{}&29&12\\
11&31&{}&{}\\
12&{}&37&6\\
13&{}&41&9\\
14&43&{}&{}\\
15&47&{}&{}\\
16&{}&53&23
\end{array}[/tex]


example the for the 10th prime =29 [tex] q= (12^2+1)/29 = 5[/tex]
and 29-12 = 17 [tex] q =(17^2+1)/29 =10[/tex]
and 29+12 = 41 [tex] q =(41^2+1)/29 =58[/tex]
and 2x29-12=46 [tex] q =(46^2+1)/29 =73[/tex]
and 2x29+12=46 [tex] q =(70^2+1)/29 =169[/tex] which is a perfect square.
etc

A further interesting property is that for many (if not all)[tex]p_2[/tex] a prime in Group 2 a infinite number of [tex]a[/tex] exists, such that [tex]\frac{(a\times p_2 \pm b)^2+1}{p_2}}[/tex] is a perfect square. (read [tex]\pm[/tex] as plus or minus b)
47318x29-12=1372210 [tex] q =(1372210^2+1)/29 =64929664969 = 254813^2 [/tex]

My question is - are there other properties that can be attributed to the Group1 or Group2 primes?

Hurkyl

Have you tried looking at your answers modulo some number? mod 2, mod 3, mod 4, or mod 8 often tell you something interesting about integer equations involving squares.

AntonVrba

Interesting - all Group2 primes have remainder 1 when divided by 4

shmoe

Except 2.

You can say more and assert that every prime congruent to 1 mod 4 is in your group 2. You're just asking for what primes p does the equation [tex]n^2\equiv -1\ \mod\ p[/tex] have a solution n, or when is -1 a square mod p. Look up the Legendre symbol, quadratic residues,Euler's criteria etc.

ramsey2879

Let a(n) = n^2 +1. Let p, q be primes from group 2 and P, Q be the unique numbers less than p/2 or q/2, respectively, such that a(P) equals 0 mod p and a(Q) equals 0 mod q.
A. If a(n) equals 0 mod p then n equals either +/- P mod p.
B. If a(P) is composite, i.e. = p*d*q (q is prime, d >/= 1) then all other prime factors of a(P) correspond to still smaller numbers Q such that a(Q) equals 0 mod q. An example is a(12) equals 0 mod 29. Since 12 < 29/2, then 12 is the lowest positive number n such that a(n) = 0 mod 29. The other prime factor of a(12) is 5 which corresponds to q = 5 where Q=2 and 12= 2 mod 5.
C. The P, Q numbers etc and the corrresponding primes (->1 means that all prime factors were previously listed) for n < 16 are
1->2
2->5
3->1
4->17
5->13
6->37
7->1
8->1
9->41
10->101
11->61
12->29
13->1
14->197
15->113

AntonVrba

Ramsey - 100% correct - this helps me further