VA !!Nf9AmQNR7I ID: 5ebae7 May 15, 2018, 7:57 p.m. No.6023   🗄️.is 🔗kun   >>6025 >>6027

>>6022

For PMA's diagram, the math is as follows:

1+ (f-1) + (n-1) = 169

1+ 133 + 35 = 169

 

The earlier n0 value would have been n0 = 10

1 + 133+ 10 = 144 = SQRT(144) = 12

 

The next value would be n0 = 62

1 + 133 + 64 = 196 = SQRT(196) = 14

 

Not sure if this works. Just brainstorming over here and keeping my mind locked on the happiness of finding the correct path! All comments and corrections appreciated.

VA !!Nf9AmQNR7I ID: 5ebae7 May 22, 2018, 11:57 p.m. No.6092   🗄️.is 🔗kun

>>6091

Hey GA! Yes, the mods come from (f-1), (n-1), (2d-1)(n-1), and nn. Here's my update. Trying to find a way to fill the x+n square with only multiples of (f-1) + mods. Last diagram is incomplete, running out of gas over here. Bed time, but back tomorrow.

VA !!Nf9AmQNR7I ID: 5ebae7 May 24, 2018, 11:13 a.m. No.6128   🗄️.is 🔗kun   >>6129

>>6127

>>6114

Thanks GA. Also, I noticed your examples are both odd and even for (x+n) squares. Since we've been working on odd squares, just wanted to let everyone know.

 

It seems like a +1 offset is needed to reach the correct x+n square in the odd examples. Maybe adding back in the -1 from (n-1)???

VA !!Nf9AmQNR7I ID: 5ebae7 May 28, 2018, 5:26 p.m. No.6191   🗄️.is 🔗kun

>>6185

Awww Yeah! Algebra Anons be like "Wut?!?!" This sounds like fun!

>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.

VA !!Nf9AmQNR7I ID: 5ebae7 June 12, 2018, 6:53 p.m. No.6334   🗄️.is 🔗kun

>>6324

Great work 3DAnon!

 

>>6325

Great work PMA!

 

>>6327

Howdy MM! Nice to see you :)

 

>>6328

>>6329

Hello VQC! As always, nice to see you. Thanks for the hints! Are the polite triangle numbers derived from (f-1) div 8 and (f-1) mod 8 the key to one of the other solutions? You hinted at that on /qr/ so just wondering.

 

>>6330

>>6333

Love this JA, Topol! Saved!

Nice get with trip 3's! Checked.