>>7193

Looks like you ended up with an extra factor of 2 starting with the line:

c+1 = (a-1)2(b-1) + a+b

Without that, the algebra does check out.

>>7202

The "((a-1)/2)(b-1)" part is just one term, so when you multiply it by 2, you should end up with only "(a-1)(b-1)".

>>7211

Unfortunately, while sqrt(p*q) = sqrt(p) * sqrt(q), the same doesn't work for sqrt(p+q) or sqrt(p-q).

>>7213

Well, I mean, it's algebraically possible to start from the (x+n)^2 equation and rearrange it to solve for sqrt(2d), but the result isn't useful since it still depends on x, n, and f.

Both d and e are easily calculated directly from c, so I'm not sure what good isolating sqrt(2d) or sqrt(e) does.

>>7215

Sure, let me do the algebra, though again I warn you that I don't think the result is particularly elegant or useful. We start with the (x+n)^2 equation you had:

(x+n)(x+n) = 2d(n-1) + nn + f – 1

x^2 + 2nx + n^2 = 2d(n-1) + n^2 + f - 1

x^2 + 2nx - f + 1 = 2d(n-1)

(x^2 + 2nx - f + 1) / (n-1) = 2d

sqrt( (x^2 + 2nx - f + 1) / (n-1) ) = sqrt(2d)

That's what solving for sqrt(2d) gives you.

A quick check with the c=559 record:

sqrt( (10^2 + 2510 - 17 + 1) / (5-1) ) = sqrt(2*23)

sqrt( (100 + 100 - 16) / 4 ) = sqrt(46)

sqrt( 184 / 4 ) = sqrt(46)

sqrt(46) = sqrt(46)