Even faster: The equation can be written as 3x + 2y = 73 - 5z. Given any value of z, based on 3*1 + 2*(-1) = 1, we get (x, y) = (73-5z)*(1, -1) + (2w, -3w), i.e. (x, y, z) = (73-5z+2w, 5z-73-3w, z), where z and w run over all integers.
Some theory: A Diophantine equation of form ax + by = c can be solved by first finding one particular solution and then adding the solutions of the homogeneous equation, just as when you solve linear differential equations. If you can find x1, y1 such that a x1 + b y1 = 1 then a particular solution is given by (xp, yp) = (c x1, c y1). The solutions of the homogeneous equation are given by (xh, yh) = (bn, -an) for all integers n. Thus the complete set of solutions are given by (x, y) = (xp, yp) + (xh, yh) = (c x1 + bn, c y1 - an).
I did this for mod 2 and mod 5 and got different equations. But then I realized that, since k and m are any integers, then my constants (call them k* and m*) could be linear combinations of k and m. And they are, it turns out that k*=k and m*=k-m+2. So either approach gives you the same (X,Y,Z) solutions.
@@SyberMath wait a minute at 3:58 why did you only write one 2 as negstive 1 you forgot the 2 in front of the z?? It doesn't make sense ti leave the other 2 like thst when you can rewrtie it as negative 1 also
Please note that the solutions are spread on that plain in such a way that they are on parellel lines in that plain, starting at point 3k-11 , -2k+23 , -k+12 on the vector 1 , 1 , -1 (choose any k for a starting point), OR starting at -m-11 , -m+23 , m+12 on the vector 3 , -2 , -1 . One line for example is (-11 , 23 , 12 ) +t(3 , -2 , -1). Another one is (-11 , 23 , 12 ) +p(1 , 1 , -1)
Here one is clear that if x is even then z is odd. x is odd then y is even.put y =x+a & z=x+b & solve. (x,y,z)=(1,10,10),(2,21,5) & many such solutions are found.
Hi ; we can write 3(x+z)+2(y+z)=73 ; we notice that : x+z odd ; so : x+z=2k+1 ( k in Z) then : y+z=35-3k Conclusion : S={(2k+1-k’; 35-3k-k’; k’)/ (k;k’)€Z^2}
Thanks. That's ok! Once you get the general idea and practice, it will become easier. Starting with two variables helps you to understand the general principles better.
I factored both sides one variable at a time by a method I'd used for other similar problems, and got x,y,z equal, respectively, 1, 5, and 12, which works. I have to admit I need to study up on modular arithmetic. Why, though, didn't modular arithmetic yield my same solution set? Again, I'm not familiar with modular arithmetic so I find it perplexing.
Without getting into all this headache, there can be a solution like.... X = 8, y= 2, z = 5...Sorry sir, I studied maths only up to 10th standard, so all your great explanation is far beyond my understanding.... But you are really a great teacher of mathematics... I salute you.. 👍👍👍✌✌
You need to add a T-shirt to your merch with some kind of half-complete mathematical process on it and the caption "Now what is that supposed to mean?"
@@joshuadorsam4619consider x-2y=1 on a 2d plane, the gradient is 1/2, solutions are at (3,1), (5,2), i.e. solutions are of the form (2n+1,n) for n in Z, i suppose something similar could be done in 3d?
If x, y, and z ∈ ℤ⁺, this system has 77 solutions.
Given x = -11 + 3k - m, y = 23 - 2k - m, and z = 12 - k + m, solutions lie in the region bound by 3k - m > 11, 2k + m < 23, and k - m < 12. There are 77 lattice points (k, m) in the interior region, yielding:
3x + 2y + 5z = 73 We can easily construct one solution: (x, y, z) = (1, 0, 14). Now, if (x', y', z') is another solution then 3(x-x') + 2(y-y') + 5(z-z') = 0. So let's solve 3u + 2v + 5w = 0. For a given value of w, we then should have 3u + 2v = -5w. We can also here easily construct one solution: (u, v) = (-5w, 5w), working for any value of w. If (u', v') is another solution then 3(u-u') + 2(v-v') = 0. So we shall now look at 3p + 2q = 0. All solutions to this are given by (p, q) = (2n, -3n). This gives (u, v) = (-5w+2n, 5w-3n) and then (x, y, z) = (1-5w+2n, 5w-3n, 14+w). Is this equivalent to (x, y, z) = (3k-m-11, 23-2k-m, 12-k+m) ? Yes, by (w, n) = (m-k-2, 2m-k-11), or reversely, (k, m) = (n-2w-11, n-w+9).
Spectacular! You can also proceed as follows after getting 3u+2v=-5w Let u=5a, v=5b, and w=-3a-2b 1-x'=u, -y'=v, 14-z'=w x'=1-u=1-5a y'=-v=-5b z'=14-w=14+3a+2b (1-5a,-5b,14+3a+2b) also gives us the general solutions
@@SyberMath 3u + 2v = -5w has (u, v, w) = (1, 1, -1) as a solution. You don't get that solution with (u, v, w) = (5a, 5b, -3a-2b) unless you take a, b rational.
I have an easier way with a more comprehensive solution, and the proof is that solution: X= -13+2k+n Y=21-3k+n Z=14-n I am looking forward to a university scholarship.. Greetings
ruclips.net/video/T_aYWFOrTmk/видео.html Very nice video SM. Bouncing off of your lead, we made a similar video, but we used basic properties of numbers to come up with other integer solutions.
Even faster:
The equation can be written as 3x + 2y = 73 - 5z. Given any value of z, based on 3*1 + 2*(-1) = 1, we get (x, y) = (73-5z)*(1, -1) + (2w, -3w), i.e.
(x, y, z) = (73-5z+2w, 5z-73-3w, z), where z and w run over all integers.
Some theory:
A Diophantine equation of form ax + by = c can be solved by first finding one particular solution and then adding the solutions of the homogeneous equation, just as when you solve linear differential equations. If you can find x1, y1 such that a x1 + b y1 = 1 then a particular solution is given by (xp, yp) = (c x1, c y1). The solutions of the homogeneous equation are given by (xh, yh) = (bn, -an) for all integers n. Thus the complete set of solutions are given by (x, y) = (xp, yp) + (xh, yh) = (c x1 + bn, c y1 - an).
I did this for mod 2 and mod 5 and got different equations. But then I realized that, since k and m are any integers, then my constants (call them k* and m*) could be linear combinations of k and m. And they are, it turns out that k*=k and m*=k-m+2. So either approach gives you the same (X,Y,Z) solutions.
Crossing my fingers for no premiere issues today
and nice thumbnail
Thanks!
everything went well with this one, excellent presentation
Thank you! 💖
@@SyberMath At 2:03 but if 5z is an even multiple of 5 then its 0 mod 2 not 1mod 2
@@SyberMath did i misunderstand?
@@SyberMath wait a minute at 3:58 why did you only write one 2 as negstive 1 you forgot the 2 in front of the z?? It doesn't make sense ti leave the other 2 like thst when you can rewrtie it as negative 1 also
Another great explanation, SyberMath! I actually found multiple values of x,y, and z.
Excellent!
Please note that the solutions are spread on that plain in such a way that they are on parellel lines in that plain, starting at point 3k-11 , -2k+23 , -k+12 on the vector 1 , 1 , -1 (choose any k for a starting point), OR starting at -m-11 , -m+23 , m+12 on the vector 3 , -2 , -1 . One line for example is (-11 , 23 , 12 ) +t(3 , -2 , -1). Another one is (-11 , 23 , 12 ) +p(1 , 1 , -1)
A great way to look at it!
Thanks for being a member!!??!?!!!!!
Here one is clear that if x is even then z is odd. x is odd then y is even.put y =x+a & z=x+b & solve.
(x,y,z)=(1,10,10),(2,21,5) & many such solutions are found.
(22,1,1) is a positive solution I found when searching for the boundaries of the positive solutions
Nice!
What is this "mod" in mathematics, can you explain me that with a video, if you want sure?!
Got some insight into solving such equations. Nice one, thanks!
Glad it helped!
Hi ; we can write 3(x+z)+2(y+z)=73 ; we notice that : x+z odd ; so : x+z=2k+1 ( k in Z) then : y+z=35-3k
Conclusion : S={(2k+1-k’; 35-3k-k’; k’)/ (k;k’)€Z^2}
I never thought we could use mod
Thanks!!
No problem!
I'm not sure why but I picture Gru teaching me maths whenever I watch these videos
😁
very helpful
What if there are four variables in diophantine equation?
Parametric solution of equation with 3 variables, cool ✌️✌️Unfortunately, I am not so fluent in handling with mods....
Thanks. That's ok! Once you get the general idea and practice, it will become easier. Starting with two variables helps you to understand the general principles better.
I factored both sides one variable at a time by a method I'd used for other similar problems, and got x,y,z equal, respectively, 1, 5, and 12, which works. I have to admit I need to study up on modular arithmetic. Why, though, didn't modular arithmetic yield my same solution set? Again, I'm not familiar with modular arithmetic so I find it perplexing.
it should
Without getting into all this headache, there can be a solution like.... X = 8, y= 2, z = 5...Sorry sir, I studied maths only up to 10th standard, so all your great explanation is far beyond my understanding.... But you are really a great teacher of mathematics... I salute you.. 👍👍👍✌✌
Thank you! 💖
An interesting fact I noticed when seeing the thumbnail
When looking for *positive integer* solutions you can easily prove that 13≥z>0
Also 22≥x>0, 32≥y>0
Can the same approach be used to solve any linear Diophantine equation? Are the mods chosen arbitrarily?
It can be. I try to make some variables disappear so choosing 2 and 3 are not arbitrary
You need to add a T-shirt to your merch with some kind of half-complete mathematical process on it and the caption "Now what is that supposed to mean?"
Good idea! 😁
The best RUclips channel
Thank you! 💖
@@SyberMath the best yt channel
Hey there! Nice work! I am back haha nice explanation btw! We can work some of the questions together!
Hello! Long time, no see! What's up? 😁
@@SyberMath Great haha wow! you have been covering a lot so far! Maybe we can work on a collab video together soon haha
Good to have you back! Sounds like a plan! 😁
@@SyberMath Hey, Sybermath! I sent you an email about it. Please check your email and get back to me :)
My solution is x equals 8-a-b and y equals 4a-b+7 and z equals b-a+7 which is a and b are any numbers
So much I like this
Thank you
You're welcome
How many different solutions?
As this equation ressembles a that of a plane in 3D space, would it be possible to solve this using tools of 3D geometry?
It would just be a line, we want to know integer solutions
@@joshuadorsam4619consider x-2y=1 on a 2d plane, the gradient is 1/2, solutions are at (3,1), (5,2), i.e. solutions are of the form (2n+1,n) for n in Z, i suppose something similar could be done in 3d?
Beautiful !!!
Thank you!
If x, y, and z ∈ ℤ⁺, this system has 77 solutions.
Given x = -11 + 3k - m, y = 23 - 2k - m, and z = 12 - k + m,
solutions lie in the region bound by 3k - m > 11, 2k + m < 23, and k - m < 12.
There are 77 lattice points (k, m) in the interior region, yielding:
(x, y, z) ∈
{(1, 5, 12), (1, 10, 10), (1, 15, 8), (1, 20, 6), (1, 25, 4), (1, 30, 2),
(2, 1, 13), (2, 6, 11), (2, 11, 9), (2, 16, 7), (2, 21, 5), (2, 26, 3), (2, 31, 1),
(3, 2, 12), (3, 7, 10), (3, 12, 8), (3, 17, 6), (3, 22, 4), (3, 27, 2),
(4, 3, 11), (4, 8, 9), (4, 13, 7), (4, 18, 5), (4, 23, 3), (4, 28, 1),
(5, 4, 10), (5, 9, 8), (5, 14, 6), (5, 19, 4), (5, 24, 2),
(6, 5, 9), (6, 10, 7), (6, 15, 5), (6, 20, 3), (6, 25, 1),
(7, 1, 10), (7, 6, 8), (7, 11, 6), (7, 16, 4), (7, 21, 2),
(8, 2, 9), (8, 7, 7), (8, 12, 5), (8, 17, 3), (8, 22, 1),
(9, 3, 8), (9, 8, 6), (9, 13, 4), (9, 18, 2),
(10, 4, 7), (10, 9, 5), (10, 14, 3), (10, 19, 1),
(11, 5, 6), (11, 10, 4), (11, 15, 2),
(12, 1, 7), (12, 6, 5), (12, 11, 3), (12, 16, 1),
(13, 2, 6), (13, 7, 4), (13, 12, 2),
(14, 3, 5), (14, 8, 3), (14, 13, 1),
(15, 4, 4), (15, 9, 2),
(16, 5, 3), (16, 10, 1),
(17, 1, 4), (17, 6, 2),
(18, 2, 3), (18, 7, 1),
(19, 3, 2),
(20, 4, 1),
(22, 1, 1)}.
Superb
Thanks 🤗
I love this
Excuse me , But how about in 4 variables?
Good use of mod😀
Thank you!
Great
3x + 2y + 5z = 73
We can easily construct one solution: (x, y, z) = (1, 0, 14).
Now, if (x', y', z') is another solution then 3(x-x') + 2(y-y') + 5(z-z') = 0.
So let's solve 3u + 2v + 5w = 0.
For a given value of w, we then should have 3u + 2v = -5w.
We can also here easily construct one solution: (u, v) = (-5w, 5w), working for any value of w.
If (u', v') is another solution then 3(u-u') + 2(v-v') = 0.
So we shall now look at 3p + 2q = 0.
All solutions to this are given by (p, q) = (2n, -3n).
This gives (u, v) = (-5w+2n, 5w-3n) and then (x, y, z) = (1-5w+2n, 5w-3n, 14+w).
Is this equivalent to (x, y, z) = (3k-m-11, 23-2k-m, 12-k+m) ?
Yes, by (w, n) = (m-k-2, 2m-k-11), or reversely, (k, m) = (n-2w-11, n-w+9).
Spectacular!
You can also proceed as follows after getting 3u+2v=-5w
Let u=5a, v=5b, and w=-3a-2b
1-x'=u, -y'=v, 14-z'=w
x'=1-u=1-5a
y'=-v=-5b
z'=14-w=14+3a+2b
(1-5a,-5b,14+3a+2b) also gives us the general solutions
@@SyberMath 3u + 2v = -5w has (u, v, w) = (1, 1, -1) as a solution. You don't get that solution with (u, v, w) = (5a, 5b, -3a-2b) unless you take a, b rational.
@@md2perpe oh yeah! That's right. Thanks. You're very good! 🤩
good job bro
Thanks
Woww cool !!!
Thanks, Tony!
X=1 y=0 z=35
I have an easier way with a more comprehensive solution, and the proof is that solution:
X= -13+2k+n
Y=21-3k+n
Z=14-n
I am looking forward to a university scholarship.. Greetings
1, 5, 12 is a set of positive solns... :) for k = m = ... guess what ;)
Cheguei cedo hoje xd
Eu tbm
Legais!
1, 5, 12
That works!
Z=12
Y=5
X=1.
Z=10
X=5
Y=4
X=6
Y=25
Z=1
X=1
Y=0
Z=14
Please read your = 12 and not 2
Put Arabic translation
I don't speak Arabic
@@SyberMath hahaha translate it
ruclips.net/video/T_aYWFOrTmk/видео.html
Very nice video SM. Bouncing off of your lead, we made a similar video, but we used basic properties of numbers to come up with other integer solutions.
You talk too fast, be slowing down a bit.