A Linear Diophantine Equation

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#NumberTheoryProblems #DiophantineEquations

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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.

mdperpe
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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.

dandjr
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Crossing my fingers for no premiere issues today
and nice thumbnail

MathElite
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everything went well with this one, excellent presentation

math
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Another great explanation, SyberMath! I actually found multiple values of x, y, and z.

carloshuertas
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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.

-basicmaths
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(22, 1, 1) is a positive solution I found when searching for the boundaries of the positive solutions

ARKGAMING
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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)

mxsjncv
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Got some insight into solving such equations. Nice one, thanks!

hsjkdsgd
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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.. 👍👍👍✌✌

ashoksatija
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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

hassanshahbou
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I never thought we could use mod
Thanks!!

nexustyrant
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Parametric solution of equation with 3 variables, cool ✌️✌️Unfortunately, I am not so fluent in handling with mods....

markobavdek
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What if there are four variables in diophantine equation?

DilipKumar-nskl
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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.

paulnokleberg
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An interesting fact I noticed when seeing the thumbnail
When looking for *positive integer* solutions you can easily prove that 13≥z>0

ARKGAMING
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Hey there! Nice work! I am back haha nice explanation btw! We can work some of the questions together!

pkmath
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I'm not sure why but I picture Gru teaching me maths whenever I watch these videos

harley_
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Can the same approach be used to solve any linear Diophantine equation? Are the mods chosen arbitrarily?

woodchuk
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What is this "mod" in mathematics, can you explain me that with a video, if you want sure?!

klementhajrullaj