/sci/ - Science & Math » Thread #10855621

2KiB, 332x224, math.png

View Same Google iqdb SauceNAO

Finding integer solutions

Anonymous Thu 01 Aug 21:35:24 2019 No.10855621 View Reply Original Report

Quoted By: >>10855670 >>10855860 >>10856469 >>10856473 >>10856485 >>10856954 >>10858838 >>10858890 >>10858977 >>10860276 >>10862926 >>10863064

Can anyone give me any sort of strategy, or source to find one, for finding integer solutions?

Like in the case of pic related, how would you find an integer solution, like (3, 4, 13, 12), without guess and check?

Anonymous

Anonymous Thu 01 Aug 2019 21:53:01 No.10855670 Report

Quoted By: >>10856485

>>10855621
hmm...not sure. but Here are a few observations

if (a,,b, c , d) is a solution, then so is (Aa,Ab, Ac,Ad) for any number A.

and if you permute a and b, the resulting will also be a solution, if you permute c and d, then the result is not a solution.

And the variety is subset of R^4, a 3 dimensional hyper surface projecting out of the origin.

Anonymous

Anonymous Thu 01 Aug 2019 22:41:20 No.10855811 Report

Quoted By: >>10856485

bump

Anonymous

Anonymous Thu 01 Aug 2019 22:52:58 No.10855841 Report

Quoted By: >>10856441 >>10856485

If a^2+b^2 is divisible by 2 but not 4, then there are no solutions. Otherwise there are finitely many solutions.

Anonymous

Anonymous Thu 01 Aug 2019 22:59:05 No.10855860 Report

Quoted By: >>10856485 >>10856500

>>10855621
Just find all the pythagorean triples and then find all squares which subtract (your c^2 - d^2 term) to get the traditional "C^2" term.
3b1 has a vid describing the way to find all possible pythagorean triples

Anonymous

Anonymous Fri 02 Aug 2019 03:51:01 No.10856441 Report

Quoted By: >>10856485

>>10855841
where did you get that from? why?

Anonymous

Anonymous Fri 02 Aug 2019 04:13:56 No.10856469 Report

Quoted By: >>10856485

>>10855621
a = b = c = d = 0

Anonymous

Anonymous Fri 02 Aug 2019 04:18:52 No.10856473 Report

Quoted By: >>10856485

>>10855621
NP problem.

if you solve it put my post number in the thesis, as a cameo.

plz

Anonymous

View Same Google iqdb SauceNAO 5-ways-your-brain-is-playing-min (...).jpg, 179KiB, 926x695

Anonymous Fri 02 Aug 2019 04:30:52 No.10856485 Report

Quoted By: >>10856491

>>10855670
linear algebra mind

>>10855811
answer seeker mind

>>10855841
number theory mind

>>10855860
geometry mind

>>10856441
curious-child mind

>>10856469
mega brainlet mind

>>10856473
complexity theory mind

and last but not least...

>>10855621
Faggot mind

Anonymous

Anonymous Fri 02 Aug 2019 04:33:41 No.10856491 Report

Quoted By: >>10858859 >>10858862 >>10858972

>>10856485
why the homophobia?

Anonymous

Anonymous Fri 02 Aug 2019 04:38:28 No.10856500 Report

Quoted By: >>10858979

>>10855860
Building off of this, take a known pythagorean triple, such as 5 (25 = 16 + 9), and build another pythagorean triple using that as one side and you have an answer. So, the next highest is 13, so you'd have to build another pythagorean triple using 13 as one side. Don't know how to avoid guess and check, but this mitigates it at least

Also, (-3,4,13,12) (3,-4,13,12) (3,4,-13,12) etc. are alternative answers ;)

Anonymous

Anonymous Fri 02 Aug 2019 09:55:43 No.10856954 Report

Quoted By: >>10860377

>>10855621
if you move d^2 to the left side, this is just a Pythagorean quadruple https://en.m.wikipedia.org/wiki/Pythagorean_quadruple

Anonymous

View Same Google iqdb SauceNAO ff.png, 19KiB, 1469x767

Anonymous Fri 02 Aug 2019 16:46:08 No.10857757 Report

Quoted By: >>10857789

No solutions except for everything being 0.

Anonymous

Anonymous Fri 02 Aug 2019 16:56:41 No.10857789 Report

Quoted By:

>>10857757
I think that just proves there's no solutions that form a valid triangle where c and d are sides of a triangle and a+b is another side.

Anonymous

Anonymous Fri 02 Aug 2019 23:05:22 No.10858838 Report

Quoted By:

>>10855621
Hi OP this is similar to pythagorean triples.
Gauss had an interesting method for finding those.
Basically you will want to use complex numbers.
(You should be familar with this by now)
And the idea is that instead of allowing all comblex numbers he chose a subset of the complex numbers called the Gaussian integers.

Basically these are the complex numbers of the form a+bi where a and b are elements of the integers.

There are lots of other methods for computing pythagorean triples aswell maybe you can adept one of these to your specifi problem

https://en.wikipedia.org/wiki/Pythagorean_triple#Relation_to_Gaussian_integers

Anonymous

Anonymous Fri 02 Aug 2019 23:12:33 No.10858859 Report

Quoted By:

>>10856491
Where do you think you are, faggot?

Anonymous

Anonymous Fri 02 Aug 2019 23:14:18 No.10858862 Report

Quoted By:

>>10856491

Leave, you don't belong.

Anonymous

Anonymous Fri 02 Aug 2019 23:22:46 No.10858890 Report

Quoted By:

>>10855621
like another anon said, this is not doable in polynomial time.
Do you know how to program? all of these are doable via program
but if you are looking for some general easy algorithm it dosen't exist

Anonymous

Anonymous Fri 02 Aug 2019 23:47:23 No.10858972 Report

Quoted By:

>>10856491
Because OP always defaults to faggot status. That's just by convention.

Anonymous

Anonymous Fri 02 Aug 2019 23:48:10 No.10858977 Report

Quoted By:

>>10855621
You could write a code to solve this in less than 5 minutes.

Anonymous

Anonymous Fri 02 Aug 2019 23:49:06 No.10858979 Report

Quoted By:

>>10856500
And the k multiples of these answers

Anonymous

Anonymous Sat 03 Aug 2019 12:15:41 No.10860276 Report

Quoted By: >>10860377

>>10855621
It should have taken you a minute to realize this is equivalent to finding rational points on the unit sphere. Now go and google that

Anonymous

Anonymous Sat 03 Aug 2019 13:18:39 No.10860377 Report

Quoted By:

>>10856954
>>10860276
Good answers. Draw a rational line through the North Pole of s sphere. Find where it intersects the sphere again. That’s a rational point. Clear denominators.

Anonymous

View Same Google iqdb SauceNAO cmMu4MI.png, 610KiB, 805x720

Anonymous Sat 03 Aug 2019 20:29:50 No.10861386 Report

Quoted By:

take any k, l
a = 2k
b = 2l + 1
c = 2k^2 + 2l^2 + 2l + 1
d = 2k^2 + 2l^2 + 2l

Anonymous

Anonymous Sat 03 Aug 2019 23:38:01 No.10861749 Report

Quoted By:

import random

while True:

a = random.randint(1,10)**2
b = random.randint(1,10)**2
c = random.randint(1,10)**2
d = random.randint(1,10)**2

ab = (a+b)
cd = (c-d)

print(a,b,c,d,ab,cd)

if ab == cd:
print(f"\nSolved: a={a**(1/2)} b={b**(1/2)} c={c**(1/2)} d={d**(1/2)}")
input()

an inefficient way

Anonymous

Anonymous Sun 04 Aug 2019 13:01:03 No.10862926 Report

Quoted By:

>>10855621
>strategy
Start with the simple approach.
All zeros is the trivial solution but it nevertheless is a solution.
One zero:
if d=0 you have a pythagorean, easy
if c=0 you have a problem
if a or b is zero you again have a pythagorean
Two zeros, and just do the permutations.

Once you have done that you will probably have a feel for how to proceed from there.

Anonymous

Anonymous Sun 04 Aug 2019 14:49:11 No.10863064 Report

Quoted By:

>>10855621
If you just want to generate integer solutions, then use the identity (w^2 + x^2)(y^2 + z^2) = (xy + wz)^2 + (wy - xz)^2, and just factorize c^2 - d^2 into c+d and c-d, and set c+d = w^2 + x^2, and c-d = y^2 + z^2 (so, a = xy + wz, and b = wy - xz).

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches