/sci/ - Science & Math » Thread #11144390

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Anonymous Thu 14 Nov 01:15:08 2019 No.11144390 View Reply Original Report

Quoted By: >>11144786 >>11144860 >>11144873 >>11144882 >>11144891 >>11146003 >>11146543 >>11146547 >>11146565 >>11147641 >>11147665 >>11147716 >>11147721 >>11148080 >>11148161

How do you solve cubic diophantine equations?

I was working on this problem

Anonymous

Anonymous Thu 14 Nov 2019 04:13:07 No.11144786 Report

Quoted By: >>11144873

>>11144390
I dunno.

Anonymous

Anonymous Thu 14 Nov 2019 04:14:51 No.11144792 Report

Quoted By:

use newton's method i dunno lol

Anonymous

Anonymous Thu 14 Nov 2019 04:47:03 No.11144860 Report

Quoted By:

>>11144390
Mind showing your work? My instinct is to start considering the quantities n^2 - 23 - m^3 (mod p) for primes p, but it's been a while since I took number theory.

Anonymous

Anonymous Thu 14 Nov 2019 04:52:24 No.11144873 Report

Quoted By:

>>11144390
I don't know 2bh

>>11144786
Me neither

Anonymous

Anonymous Thu 14 Nov 2019 04:55:02 No.11144882 Report

Quoted By:

>>11144390
i don't know how to do it, sorry

Anonymous

Anonymous Thu 14 Nov 2019 04:58:15 No.11144891 Report

Quoted By:

>>11144390
Dont know, sorry op

Anonymous

Anonymous Thu 14 Nov 2019 15:37:19 No.11146003 Report

Quoted By:

>>11144390
This is probably wrong, but I always wanted to try: solve the cubic, then solve as a diophantine.

Anonymous

Anonymous Thu 14 Nov 2019 15:58:59 No.11146027 Report

Quoted By:

I tried a few hundred numbers for B none lead to a a square

Anonymous

Anonymous Thu 14 Nov 2019 19:26:20 No.11146543 Report

Quoted By:

>>11144390
bumping for the unknowns to be known

Anonymous

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Anonymous Thu 14 Nov 2019 19:27:23 No.11146547 Report

Quoted By:

>>11144390
What have you done so far?

Anonymous

Anonymous Thu 14 Nov 2019 19:36:01 No.11146562 Report

Quoted By:

I have just been looking at different modulus classes (for example (mod 23)), and I was hoping that some mod class wouldn't have residues that satisfy the equation. But haven't found anything useful so far

Anonymous

Anonymous Thu 14 Nov 2019 19:38:04 No.11146565 Report

Quoted By:

>>11144390
This looks like a job for
>Q U A N T U M C O M P U T I N G M A N

Anonymous

Anonymous Thu 14 Nov 2019 20:28:54 No.11146713 Report

Quoted By:

bumping

im playing around with it but i have no fucking clue desu

Anonymous

Anonymous Thu 14 Nov 2019 20:29:40 No.11146717 Report

Quoted By:

If i remember, ill try to solve this once i get home and have access to python

Anonymous

Anonymous Thu 14 Nov 2019 21:29:03 No.11146912 Report

Quoted By:

just google it

Anonymous

Anonymous Thu 14 Nov 2019 21:44:34 No.11146963 Report

Quoted By: >>11147095 >>11147628

Wrote a python script, can confirm no solutions up to 1<=a<= 10000000

Anonymous

Anonymous Thu 14 Nov 2019 22:39:29 No.11147095 Report

Quoted By: >>11147628

>>11146963
find(199900000,999999999)
gives no solution in A, i think we can call it a day boys

Anonymous

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Anonymous Fri 15 Nov 2019 01:40:59 No.11147628 Report

Quoted By:

>>11146963
>>11147095
>computer say no solution up to big number, must be no solution

Anonymous

Anonymous Fri 15 Nov 2019 01:45:49 No.11147641 Report

Quoted By: >>11147708

>>11144390
Well why not try to first figure out where A2 = B3

Anonymous

Anonymous Fri 15 Nov 2019 01:54:46 No.11147665 Report

Quoted By:

>>11144390
Well A = sqrt 23+b^3 as you have written.

Anonymous

Anonymous Fri 15 Nov 2019 02:16:43 No.11147708 Report

Quoted By: >>11147724

>>11147641
>A^2=B^3
Let B be any integer squared. That means B=n^2, so B^3=(n^2)^3=n^6=(n^3)^2=A^2 where A is guaranteed to be an integer.

Anonymous

Anonymous Fri 15 Nov 2019 02:20:12 No.11147716 Report

Quoted By: >>11147978

>>11144390
A=269586136, and B=417316

This can't be right, can it?

Anonymous

Anonymous Fri 15 Nov 2019 02:24:13 No.11147721 Report

Quoted By: >>11148060

>>11144390
My entire IP range is banned from posting images , so I'll just ask my question here.
How do you solve the recurrence relation : aN = 7aN-2 + 6aN-3 , where a0 = 9,a1=10 and a2=32. I tried plugging in the numbers to generate more terms but couldn't spot a pattern. I know it has something to do with generating functions, but I don't know what they are and I also don't know why they help us solvr this problem.

Anonymous

Anonymous Fri 15 Nov 2019 02:24:44 No.11147724 Report

Quoted By: >>11147737

>>11147708
Z=A^2-23 = B^3

B = cubert Z

Anonymous

Anonymous Fri 15 Nov 2019 02:29:52 No.11147737 Report

Quoted By: >>11147778

>>11147724
the whole point of this thread is to find integers that satisfy the equation, retard

Anonymous

Anonymous Fri 15 Nov 2019 02:40:49 No.11147773 Report

Quoted By: >>11147869 >>11147963 >>11148010

its a mordell curve, theres documented solutions, and 23 has no solutions. not sure how to prove it though

Anonymous

Anonymous Fri 15 Nov 2019 02:42:04 No.11147778 Report

Quoted By:

>>11147737
sorry im retarded, couldnt you tell, retard?

Anonymous

Anonymous Fri 15 Nov 2019 03:07:33 No.11147869 Report

Quoted By:

>>11147773
damn, i didnt know how to search for this. thanks math anon.

Anonymous

Anonymous Fri 15 Nov 2019 03:41:18 No.11147963 Report

Quoted By:

>>11147773
Thank you! this is super useful, now i can atleast try to think of how to prove the no solutions case

Anonymous

Anonymous Fri 15 Nov 2019 03:46:10 No.11147978 Report

Quoted By: >>11147993 >>11148001

>>11147716
This is not right im afraid, (if you plug the values into wolfram, itll show you a lot more decimals) where did you get these values from anyway?

Anonymous

Anonymous Fri 15 Nov 2019 03:51:42 No.11147993 Report

Quoted By:

>>11147978
I wrote a small program in MATLAB that just runs through integers, testing if they work. I figured they only showed up because the floating points get rounded. Any idea of how to get around this? I know now that A^2=B^3+b has no solutions for b=23, but I want to find solutions for other values of b.

Anonymous

Anonymous Fri 15 Nov 2019 03:54:47 No.11148001 Report

Quoted By:

>>11147978
here's my code

b=23;
for n=1:100000 %keep n<~100000
f=n^3+b;
if floor(sqrt(f))==ceil(sqrt(f))
fprintf('For f(n)=n^3+(%1.0f), n=%1.0f and sqrt(f(n))=%1.0f\n',b,n,sqrt(f))
return
end
end

and yes, I know I'm trash

Anonymous

Anonymous Fri 15 Nov 2019 03:59:09 No.11148010 Report

Quoted By:

I'm this dude >>11147773, and i found a paper (second google result, by Keith Conrad) that proves it for many cases. specifically look at theorem 2.3, the only difference is we'd have y2 + 4 = x3+27

Anonymous

Anonymous Fri 15 Nov 2019 04:19:13 No.11148060 Report

Quoted By: >>11148100 >>11148116

>>11147721
PLS respond

Anonymous

Anonymous Fri 15 Nov 2019 04:29:46 No.11148080 Report

Quoted By:

>>11144390
I fucking hate integers and integer masturbation.

Anonymous

Anonymous Fri 15 Nov 2019 04:36:52 No.11148100 Report

Quoted By:

>>11148060
a3=7(10)-6(9)=16
a4=7(32)-6(10)=164
and so on.
i dont know what your confusion is

Anonymous

Anonymous Fri 15 Nov 2019 04:43:12 No.11148116 Report

Quoted By: >>11148133

>>11148060
discrete openmathbooks org/dmoi2/sec_recurrence.html

Anonymous

Anonymous Fri 15 Nov 2019 04:49:39 No.11148133 Report

Quoted By:

>>11148116
however, it seems you get a cubic for the characteristic polynomial since you have a_n = ba_n-2 + ca_n-3, instead of n-1 and n-2. just some extra work

Anonymous

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Anonymous Fri 15 Nov 2019 05:00:53 No.11148161 Report

Quoted By:

>>11144390

Anonymous

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Anonymous Fri 15 Nov 2019 09:46:12 No.11148647 Report

Quoted By:

here u go OP
b=1, a = 1

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