Anonymous ID: 1599bb Dec. 30, 2019, 11:51 p.m. No.7671176   🗄️.is 🔗kun   >>1269

>>7671110

I have two questions about that. Firstly, when we have a narrowed-down range like that, does it actually make it any more efficient to find the number of primes below a number (since as far as I'm aware you'd still have to iterate to test them all, regardless of the range being narrowed down)? Secondly, if you do primepi(d) it tells you how many a values you'd have to search through, but how does that relate to finding the factors of c in O(log n)?

 

>>7671143

a*b=c

d=sqrt(c)

e=c-dd

f=e-(2d+1)

x=d-a

n=((a+b)/2)-d

(d+n)(d+n)-(x+n)(x+n)=c

{e:n:d:x:a:b}

Anonymous ID: 1599bb Dec. 31, 2019, 12:36 a.m. No.7671394   🗄️.is 🔗kun   >>1444

>>7671269

We don't appear to be aware of it already (it isn't in the GP thread and quick tests with knowns mod 4 haven't shown us anything). Maybe you could give us a nudge in the right direction?

Anonymous ID: 1599bb Dec. 31, 2019, 1:04 a.m. No.7671517   🗄️.is 🔗kun

>>7671444 (nice trips)

Just so we know if we understand, an example.

c559 = 13*43

13%4=1, 4*3+1=13

43%4=3, 4*10+3=43

The +1 and +3 being the distances between a and b and a multiple of 4.