dChan
Anonymous ID: 43930d Oct. 14, 2018, 2:59 p.m. No.7808   🗄️.is 🔗kun   >>7809 >>7810

An interesting relation I found while working on the new hints that could become a quicker way to calculate integer square roots:

 

TM1(c-1/2) = d or d-1

Anonymous ID: 43930d Oct. 14, 2018, 10:55 p.m. No.7831   🗄️.is 🔗kun   >>7833 >>7841

c6107

{23:2976:78:77:1:6107} (23, 2976, 39)

{23:1:18641629:-6107:18647736:18635524} (23, 1, -3053)

{23:1:18653843:6107:18647736:18659952} (23, 1, 3054)

 

{-134:1:18647657:-6108:18653765:18641551} (-134, 1, -3053)

{-134:1:18659873:6108:18653765:18665983} (-134, 1, 3055)

 

α = 18641607

β = 9322268

γ = 18647657

δ = 18647712

 

γ - α = 6050

δ - γ = 55

γ - α - c = -57

 

α - (-134, 1, -3053)[b] = 56

α - (-134, 1, -3053)[a] = -12158

(α - (-134, 1, -3053)[a])/2 = -6079

α - (23, 1, -3053)[b] = 6083

α - (23, 1, -3053)[a] = -6129

α - (23, 1, -3053)[b] - α - (23, 1, -3053)[a] = -46

 

6107[x] = 47

Anonymous ID: 43930d Oct. 14, 2018, 10:58 p.m. No.7832   🗄️.is 🔗kun   >>7841

c34117

{261:16875:184:183:1:34117} (261, 16875, 92)

{261:1:581950858:-34117:581984975:581916743} (261, 1, -17058)

{261:1:582019092:34117:581984975:582053211} (261, 1, 17059)

 

{-108:1:581984790:-34118:582018908:581950674} (-108, 1, -17058)

{-108:1:582053026:34118:582018908:582087146} (-108, 1, 17060)

 

α = 581950598

β = 290983839

γ = 581984790

δ = 581984713

 

γ - α = 34192

δ - γ = -77

γ - α - c = 75

 

α - (-108, 1, -17058)[b] = -76

α - (-108, 1, -17058)[a] = -68310

(α - (-108, 1, -17058)[a])/2 = -34155

α - (261, 1, -17058)[b] = 33855

α - (261, 1, -17058)[a] = -34377

α - (261, 1, -17058)[b] - α - (261, 1, -17058)[a] = -522

 

34117[x] = 75

 

This hasn't been ironed out, as one can see.