Anonymous ID: e5f0f6 May 25, 2019, 2:14 a.m. No.9202   🗄️.is 🔗kun

>>9201

Yeah I think you're on to it. The more factors a product has the more (x+n) squares does it have.

 

>>9200

The only constants we have for all our (x+n)^2 squares are d and f. Everything else depends on the different records.

 

But then if h is a variable, what is it connected to? We have h-families. Is then h related to the number of factors a product has? For a semiprime, we have 2 families? I remember the >>5690

>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle.

 

Although I don't think we ever figured out how polite numbers are related to the triangles, did we? Nor do I think we ever actually "got" the configuration of the triangles correctly either.

Anonymous ID: e5f0f6 May 25, 2019, 2:33 a.m. No.9203   🗄️.is 🔗kun   >>9204

>>9200

Take c=145, it has a=5, b=29. Meaning it has a family of 4, when taking f into consideration? Unless we count a=1, b=145 in which case it has a family of 8?

Name ID: e5f0f6 May 25, 2019, 3:17 a.m. No.9205   🗄️.is 🔗kun   >>9206

If I'm correct. For RSA2048:

(n-1)(n-1) - 1 mod 8= 0

2n mod 8 = 6

2(n-1)d mod 8 = 0

f-2 mod 8 = 0

 

This we know 2n has to be on the form 8k + 6 which means n = 4k + 3 for some k.