Anonymous ID: cb5395 Dec. 21, 2017, 5:25 a.m. No.854   🗄️.is 🔗kun   >>856

Problem is to solve for n where a and b are prime. in record (0,2,c^2,x,a^2,b^2) you can solve for n in terms of a and b.

x=a^2(b^2-1)

2an=x^2

n=a^2(b^2-1)^2

Anonymous ID: cb5395 Dec. 21, 2017, 5:33 a.m. No.856   🗄️.is 🔗kun

>>854

Sorry again

record is (0,n(c^2),c^2,x,a^2,b^2) so you can solve n for squared record in terms of square of original primes.

n(c^2)= a^2(b^2-1)/2

Anonymous ID: cb5395 Dec. 21, 2017, 5:45 a.m. No.857   🗄️.is 🔗kun

To be clearer

For record (0,n(c^2),c,x(c^2),a^2,b^2)

for primes a,b

x(c^2)^2=2a^2n(c^2)

x(c^2)=a^2(b^2-1)

n(c^2) =a^2(b^2-1)/2

using x(c^2) and n(c^2) to distinguish from x and n for record (e,n,c,x,a,b)