Solving a Diophantine System with Three Variables

This video is about solving a Diophantine System
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#NumberTheoryProblems #DiophantineEquations

EXPLORE:

[davidchung]

Another way to solve the system is to square the bottom equation and then add it to the first equation. This gives:
2X^2 - 2 x y - 2 z x + 2 y z = 10.

The above factors into

(x - y) ( x - z) = 5.

So, x - y = +/-5 or x - y = +/- 1, which lead to 4 solutions.

[tonyhaddad]

I like diophantine system its make a lot of sense i have solved by squaring the second equation and sum it with the first you get a common factor equal to 5 (prime number cool You can use the second equation to change the factors to (3-y)(3-z), there are 4 valid set of solution, and we done

[MrStudent]

replace x with y+z-3 in 1st equation....you will reach (z-3)(y-3)=5....this will provide the results as you obtained.

[srijanbhowmick]

My solution :---
y + z - x = 3
=> x = y + z - 3
x^2 - y^2 - z^2 = 1
=> (y + z - 3)^2 - y^2 - z^2 = 1
After some simplification, we get :---
y*z - 3y - 3z + 4 = 0
=>y(z - 3) - 3(z - 3) - 9 + 4 = 0
=> (y - 3)(z - 3) = 5
Now,
5 = 5 * 1
= 1 * 5
= (-5) * (-1)
= (-1) * (-5)
By casework, and x = y + z - 3, we get the following triplets :---
x y z
9 8 4
9 4 8
-3 -2 2
-3 2 -2

[elkincampos]

Use the paralelogram identity in 2:00. We can use that y=(1/2)*(y+z)+(1/2)*(y-z). But (y+z)^2+(y-z)^2=2*(y^2+z^2). Then This is a square.

[RobG]

The clever, multi-level substitutions are like a mathematical equivalent of Christopher Nolan's movie Inception.

[Qermaq]

10:45 I subtracted the two equations instead of adding them, giving me 2w = 8, w = 4, and importantly w^2 = 16. Dumped that back in the x^2 - 6x - 11 = w^2 and solved for both values of x, giving me all 4 possible solutions (as y and z are interchangeable wlog).

[thomshrike]

My solution is probably simpler; (i) the first equation is (x+y)*(x-y)=z^2+1 and you can replace (x-y) with (z-3) so that it becomes: (x+y)=(z^2+1)/(z-3); (ii) add this equation to equation #2, so that you get rid of y, simplify and get x=(z^2-3z+5)/(x-3) and then y=(3z-4)/(z-3)=3+(5/(z-3); (iii) hence, for y to be an integer, z has to be one of the following: -2, 2, 4, 8 and you have all the solutions.

[adityaprakash]

Why so difficult solution?

Express both x and y in terms of z
It can be done by isolating z on one side
And now both x and y will be in terms of (some integer + 5/(z-3))
It can be interger only when z is either 2, 4, -2, and 8

[-basicmaths]

Answer (x, y, z)=(-3, -2, 2)
(x, y, z)=(-3, 2, -2)

[leecherlarry]

no biggo with a bit of help hehe:
*Solve[{x^2 - y^2 - z^2 == 1, y + z - x == 3}, {x, y, z}, Integers]*

[elias]

Help. Every time I read "W" I read it as "Wuh", "Woah" or "Wih". And _never_ "double you"

[neuralwarp]

Your screen keeps going out of focus. Very annoying.

[neuralwarp]

Sorry: 2+ advert breaks means an automatic thumbs down from me