The alpha question became the alpha conjecture 

The alpha question. Let be a prime (greater than 3, to prevent it from being 3) such that 

(the ' -1' case) and are both prime, or  

(the '+1' case) and are both prime  

is it then true that (mod p)?
 

(Private terminology. I took to calling the other primes 'inexplicable' for the time being. Other later examples, when I extended my searching were:  25117 - 1 = () (3) (7) (13) (23),  60919 - 1 = (2) (3) (11) (13) (71), 

92401 - 1 = () (3) () (7) (11),  115827 - 1 =() (3) () (197), etc etc.) 

There were two first cases to examine, with unimportant order: with '+1' or '-1'. 

Let's say we begin, as I did, with and '-1': thus we are trying to find primes { p } made up as follows: 

, where are primes with , () 

and are hoping that such primes satisfy: 

(mod p) 

Such p's are like this: 

  =  

which is, as I observed at the time, a quadratic whose discriminant is  

Here I show and '-1' for forming some primes { p }.  

I abbreviate 'alpha' to 'a' in the following to reduce the program lines in the following, in which: 

ithprime(k) is the ithprime, 'k' starting at 3 ("from 3"): 5, 7, 11, 13, 17, ...
So, that produces the candidates for  

isprime(2^a*ithprime(k)+SIGN) produces the candidates 2^a*ithprime(k)+SIGN for  

isprime(2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1) produces the desired primes { p },
the 2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1 

>

a := 1: SIGN := -1: count:= 0: for k from 3 to 100 do if isprime(2^a*ithprime(k)+SIGN) and    isprime(2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1)                     then count := count+1: q1[count] := ithprime(k): q2[count] := 2^a*q1[count]+SIGN:  p[count] := 2^a*3*q1[count]*q2[count]+1:   fi od; array([[q[1], q[2], p], seq([q1[k], q2[k], p[k]], k=1..count)]);

 

(6.1)

 

>

 

At the time at which I tested the above, I had systematically tested for (mod p), for all primes up to the 30,000th (note to myself: all that is saved in a Maple worksheet that I had called 'Wilson play.mws', the name of which gives an indication of my attitude up until then). The largest such prime p I had found at that stage had been 347821 (an 'inexplicable' prime,   being *3*5*11*17*31), and thus those three last new primes - 627919, 879667, and 2310019 - in the above table were first tester candidates of my alpha observation... (heart beating...). 

Well, I wouldn't be here talking to you if it hadn't turned out that:
 

>	p1 := 627919: P[1,3](p1) mod p1; # didn't produce a '1'

 

(6.2)

 

>	1372^3 mod p1; # but its cube did

 

(6.3)

 

>	p2 := 879667: P[1,3](p2) mod p2;

 

(6.4)

 

>	p3 := 2310019: P[1,3](p3) mod p3;

 

(6.5)

 

>

 

The reader may imagine how I felt seeing those 1's... 

>

a := 1: SIGN := -1: count:= 0: for k from 501 to 1000 do if isprime(2^a*ithprime(k)+SIGN) and    isprime(2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1)                     then count := count+1: q1[count] := ithprime(k): q2[count] := 2^a*q1[count]+SIGN:  p[count] := 2^a*3*q1[count]*q2[count]+1:   fi od; array([[q[1], q[2], p], seq([q1[k], q2[k], p[k]], k=1..count)]);

 

(6.6)

 

>

 

Here I show and '+1' for forming some primes { p }: 

>

a := 1: SIGN := +1: count:= 0: for k from 3 to 100 do if isprime(2^a*ithprime(k)+SIGN) and    isprime(2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1)                     then count := count+1: q1[count] := ithprime(k): q2[count] := 2^a*q1[count]+SIGN:  p[count] := 2^a*3*q1[count]*q2[count]+1:   fi od; array([[q[1], q[2], p], seq([q1[k], q2[k], p[k]], k=1..count)]);

 

(6.7)

 

>

 

I tested alpha up to 10 with both '-1' and '+1' choices, and checked if (mod p) held, not for all the produced primes { p }, but only because of computation time constraints: for example, the prime 18711862657 (see below: , , ) took 80.5 hours to test on a 2.8GHz Pentium 4.  

Here I show and '+1' for forming some primes { p }: 

>

a := 7: SIGN := +1: count:= 0: for k from 3 to 200 do if isprime(2^a*ithprime(k)+SIGN) and    isprime(2^a*3*ithprime(k)*(2^a*ithprime(k)+SIGN)+1)                     then count := count+1: q1[count] := ithprime(k): q2[count] := 2^a*q1[count]+SIGN:  p[count] := 2^a*3*q1[count]*q2[count]+1:   fi od; array([[q[1], q[2], p], seq([q1[k], q2[k], p[k]], k=1..count)]);

 

(6.8)

 

>

 

Of course I knew I was onto something: 

The alpha conjecture. Let be a prime (greater than 3, to prevent it from being 3) such that 

(the ' -1' case) and are both prime, or  

(the '+1' case) and are both prime  

then (mod p)
 

But the problem then became: how can one prove this (almost certainly true) conjecture? That brings me to the next natural section: