>>4219

That sneaky VQC. I think he's just given us a new equation.

>(x+n)(x+n) = nn + 2d(n-1) + f - 1

Using distributive property, (x+n)(x+n)= xx + xn + xn + nn

So for this portion: 2d(n-1) + f - 1 = 2xn + xx

Do we have this equation already? Maybe we do, can't find it in my notes tho.

>>4251

Great visualization, Teach! Check out the one I've attached. For (x+n)=15. For odd (x+n) there's always a square 1x1 in the middle. Right triangles, and exterior side will always be one more than the interior side. Thoughts?

>>4546

Set it up! "Hobo's Math Camp"

Staying up late here. Hoping VQC will pop by. Seems like I'm always in bed bc of time difference.

This is what I was talking about the other night in a very unclear manner. If we have n0, possibly some combo of n0^2 and 2d can fill the triangle. Visual is large triangle filled with multiples on n0.

>>4647 PMA is close here! Maybe a remainder or factor to be added back in?

>>4654

This means the area, not the base.

>>4784

Ok, fair enough PMA! Just sayin the old equation for f solves for all (e,1).

>>4770

>>4768

So is this equation still solid? MA verified?

(x+n)^2 = n^2 -1 + 2d(n-1) + 2d + 1 - e

>>4808

Hello PMA! I verified your formulas and calcs for each example you’ve posted.

>>4812

Thanks MM! It will hopefully help us iterate up quickly to (x+n) using n0 triangle geometry. Thinking out loud in diagram form. Next version I’ll work in 2d multiples.

>>4343

VQC’s post about filling the odd x+n square using f-2 doubling.

>n0^2 + multiples of 2d will fill the triangle.

>>4839

>Good to see the board wasn't flooded with shitposts while I was away.

Yes, I have changed my evil ways AA. Only working now. 5:1 work to shitpost ratio ok with everyone? ;)

>>4879

I think I have something solid here, lads!! Been working on this all morning. Can I please get some eyes on this to verify?

So next question is: we're building (x+n) without knowing it's full size starting only from (f-2). How do we know when we have a match starting only from c? We need a method to verify correct (x+n), and I'm sure PMA is on it. The following equations could help create a crosscheck to verify correct (x+n) and also solve for n and x. Maybe we increase the n0 factors until we get a balanced lock on all equations? Check it out ==>>

If we know c = (d+n)^2 - (x+n)^2 and we can solve for (x+n) then we're going to be able to use the quadratic from RSA#2 to solve for n. Here's the updated version using (TuXN)^2 instead of (2t)^2.

Find TuXN = (x+n) = 15 using (f-2) method. Verify using .

Using our TuXN = 15 prime element: {3:6:16:9:7:37}

n = SQRT(c + (TuXN)^2) - d

n = SQRT(259 + (15^2)) - 16 = 6

Then plug in n to solve for x, and then a and b

x = SQRT((d+n)^2 - c) - n

x = SQRT((16+6)^2 - 259) - 6 = 9

d - x = a = 16 - 9 = 7

c/a= b = 259 / 7 = 37

Element complete starting only from C!!

>>4917

No worries, MM. Here's a rundown of what I currently know. Please correct any errors, Anons. Working to understand over here.

VQC's claim is this: because of triangle geometry, we can start with f, and use it to divide the odd (x+n) square into 8 triangles + 1.

Then we find n0, which is (f-2)/8 = n0 (whole integer) + mod (remainder).

Then we use multiples of n0 to iterate upward to find the small square (x+n) that matches c = (d+n)^2 - (x+n)^2.

We can use the formulas I've posted to check each iteration of n0*factor as True or False for a match to this formula: c = (d+n)^2 - (x+n)^2

Iterate, check True or False. Upward through all possible values of (x+n) until we get a match.

>>4922

Awesome way to find the shortcut, Baker! Have you checked it out?

I need to run another test case, lads. Have a few things to do first, but I'll have another example up soon to see if this actually works.

>>4924

In PMA's output, third example for c = 6017: notice that the correct (x+n) is not exactly found. However, the difference is so close! Also, the proximate answers could be iterated by +1 or -1 until a match is found.

>>4973

I’ll build a spreadsheet tonight to demonstrate. Gotta finish a few things first. We can get to the correct x+n quickly.

>>4989

Hello PMA! I examined the equations, and they look good. So how do we use them? Interesting that CA is also working on remainders of integers right now.

>>4998

>>4999 Checked!

Hello CA! I'm finding a similar pattern to your chart as I iterate (x+n) values upward towards the correct (x+n) value. Your idea is closely related to the number trees, right? Can you explain in more detail how to read this chart? I'm following your ideas, and understanding them.

>These are the continued fractions of the square root of each number. The highlighted part is the part that repeats and the starting element is just our d. Notice that each series ends with the value 2d. Also notice that if 2d is divisible by e (the column after the d column) Then the continued fraction is [d; (2d/e, 2d)]

>Then I noticed that they were always .25 above the prime value, and I remembered that every odd square minus one is divisible by 8 and therefore dividible by 4, and also if I subtracted the one I would get rid of that .25.

>Since this exists you could probably reverse it, like have a number and a "key" a and subtract a from the number then multiply by 4. Then maybe or maybe not add 1.

>Maybe this could be how the tree system works too.