I think you made a mistake in the first method, as to get the general solution for Z you should do (Z1 - A/gcd * S) being Z1 the particular solution for Z, A the coeficient of Y and S a random integer. Well your soultion is Z=-2-11S but it should be Z=-2-5S as 5 is the coeficient of Y in the original equation (and it is divided by gdc=1 but that does not change anything) Besides that, first method works great. thanks for sharing!
Hi, this is one of the best videos outlining, in brief, the different methods to solve a Diophantine Equation. Which book can I find questions related to "parameterizing with canonical form". I will be really grateful if you can help out because I need it for a project. Thank you for all your help in advance.
I understand this was posted over a year ago, but I'm certain that your method 1 solution is incorrect. If t=0 and s=1, you would have x=5, y=160 and z=-208, which gives a large negative number, not 1. I think it came from a typo. What you're looking for is the following: x=-5+11t y=(3+7s)(16-35t) z=(-2-5s)(16-35t)
My solution involves doing some modulo 5, 7 and 11 since they all are divisors of exact two of the numbers. Modulo 5: z = 3 => z = 5c + 3 Modulo 7: y = 6 => y = 7b + 6 Modulo 11: x = 6 => x = 11a + 6 35(11a+6) + 55(7b+6) + 77(5c+3)=1 a+b+c = -2 (a, b, c) ∈ {(a, b, -(a+b+2) | a,b ∈ Z} (x, y, z) ∈ {(11a+6, 7b+6, -5a-5b-7) | a,b ∈ Z}
@@thinkinginmath3009 Third method is standard for all linear equations , no matter they come from algebra (system of linear equations) or from calculus (systems of linear ode) but is it true or maybe i have troubles with homogeneous equation Consider following equation 15x + 9y + 7z = 97 Particular solution is easy to guess for example x0 = 0 y0 = 10 z0 = 1 But i tried to solve equation 15x + 9y + 7z = 0 and probably i solved it wrong 3(5x+3y)+7z = 0 3(5x+3y) = -7z z = 3s 3(5x+3y) = -7*3s 5x+3y = -7s What's next according to your video 5x+3y = -7s I could solve it as linear nonhomogeneous with two variables but you in your example did it differently
@Dan Hosu SyberMath likes solution which you presented I prefer third solution presented on tis video because it is standard approach for any linear equation no matter where it come from fe from discrete mathematics (recurrence relations and systems of recurrence relations) , from algebra (systems of linear equations) , from calculus (linear ode and systems of linear ode)
I think you made a mistake in the first method, as to get the general solution for Z you should do (Z1 - A/gcd * S) being Z1 the particular solution for Z, A the coeficient of Y and S a random integer. Well your soultion is Z=-2-11S but it should be Z=-2-5S as 5 is the coeficient of Y in the original equation (and it is divided by gdc=1 but that does not change anything)
Besides that, first method works great.
thanks for sharing!
indeed, it should be -2-5s instead of 11s. Thanks for pointing out the error here :)
Hi, this is one of the best videos outlining, in brief, the different methods to solve a Diophantine Equation.
Which book can I find questions related to "parameterizing with canonical form". I will be really grateful if you can help out because I need it for a project. Thank you for all your help in advance.
hello. for method 1, are the values of s and t always equal?
I understand this was posted over a year ago, but I'm certain that your method 1 solution is incorrect. If t=0 and s=1, you would have x=5, y=160 and z=-208, which gives a large negative number, not 1. I think it came from a typo. What you're looking for is the following:
x=-5+11t
y=(3+7s)(16-35t)
z=(-2-5s)(16-35t)
My solution involves doing some modulo 5, 7 and 11 since they all are divisors of exact two of the numbers.
Modulo 5: z = 3 => z = 5c + 3
Modulo 7: y = 6 => y = 7b + 6
Modulo 11: x = 6 => x = 11a + 6
35(11a+6) + 55(7b+6) + 77(5c+3)=1
a+b+c = -2
(a, b, c) ∈ {(a, b, -(a+b+2) | a,b ∈ Z}
(x, y, z) ∈ {(11a+6, 7b+6, -5a-5b-7) | a,b ∈ Z}
Excellent! Thanks for sharing!
@@thinkinginmath3009 Third method is standard for all linear equations , no matter they come from algebra (system of linear equations)
or from calculus (systems of linear ode)
but is it true or maybe i have troubles with homogeneous equation
Consider following equation
15x + 9y + 7z = 97
Particular solution is easy to guess for example
x0 = 0
y0 = 10
z0 = 1
But i tried to solve equation
15x + 9y + 7z = 0
and probably i solved it wrong
3(5x+3y)+7z = 0
3(5x+3y) = -7z
z = 3s
3(5x+3y) = -7*3s
5x+3y = -7s
What's next according to your video
5x+3y = -7s
I could solve it as linear nonhomogeneous with two variables but you in your example did it differently
@Dan Hosu SyberMath likes solution which you presented
I prefer third solution presented on tis video because it is standard approach
for any linear equation no matter where it come from fe from discrete mathematics (recurrence relations and systems of recurrence relations) ,
from algebra (systems of linear equations) , from calculus (linear ode and systems of linear ode)
Top !
please explain thoroughly, i don't understand the flow of your solution. Some part is just to0 fast, can't keep up
I see. I will plan to add another video on this, with more details.
@@thinkinginmath3009 thank you so much. Will wait till you uploud another vid
What didn't you understand?
US and NATO planes and drones fly from the Black Sea every day for 3 years. What exactly is your problem?