A System of Equations with Integer Solutions

If you need to post a picture of your solution or idea:
#NumberTheoryProblems #DiophantineSystems

EXPLORE 😎:

PLAYLISTS 🎵 :

[jasimmathsandphysics]

I got the 2021, 0, 2020 solution before watching the video I’m officially intellectual 😂

[babangmalas]

I'm very excited sir.... I have try it before see the your explation, although I only got 1 solution 0, 2020, 2021 thank you very much sirrr.... 😁

[scottleung]

Yes, I got it! I think this one was easier than most.

[italixgaming]

We have: yz+x-xy-z=z(y-1)+x(1-y)=1 <=> (z-x)(y-1)=1
Since x, y and z are integers, z-x and y-1 are integers.
The only solutions of ab=1 in Z are (1, 1) and (-1, -1) (if a=0 or b=0 the product is 0 and if |a|>1 for example |ab|>1).
If y=2, z=x+1. In the first equation we have: 2x+x+1=2020 => x=673.
Our first solution is (673, 2, 674).
If y=0, z=x-1 and the first equation becomes: x-1=2020 => x=2021.
Our second solution is (2021, 0, 2020)

[-basicmaths]

Really good problem.
Subtract first equation from second equation We get
(y-1)(z-x)=1
y-1=1
y=2, and z=x+1
Now xy+z=2020
2x+x+1=2020
3x=2019
x=673 and z=674
Recheck with second equation
x+yz=673+(2*674)
=673+1348=2021

[quantumobject]

I'll just add a comment to support your great work

[StuartSimon]

What happened to the closing brace for the solution set?

[coolmangame]

Fun fact this problem was given in a mathematical olympiad for 8th graders in my country few momths ago. Took me a little too long to solve it as a 9th grader tho lol

[voodooros]

When we're in the integer world, I guess m, n, p etc letters for variables makes more sense than x, y, z etc.

[Qermaq]

Haven't watched this yet but if I found the correct answer, boy are you evil! :D Tried everything till I found that it was all for "nothing" ;) (Edit - I only found the second, missed the first.)

[ThangNguyen-qpxd]

xy+z=2020
x+yz=2021

Because x, y, z are integers, so we have two cases.
Case 1: 1-y=1 and x-z=1<=>y=0 and x=z+1.
With y=0 then z=2020 and x=2021 (replace y=0 into the two above equations)
Case 2: 1-y=-1 and x-z=-1 <=> y=2 and z=x+1.
Replace into the first equation, we have 2x+x+1=2020<=>3x=2019<=>x=673 and z=673.
Concluding, we have two solutions (2021, 0, 2020) and (673, 2, 674)

[raystinger]

So first thing I do is to add both equations we have:
x + xy + z + yz = 4041
(x + z)(y + 1) = 4041 = 3²*449
Since (x + z) and (y +1) are both integer factors, we have the following possible values for (y + 1):
(y + 1) ∈ {1, 3, 9, 449, 1347, 4041} => y ∈ {0, 2, 8, 448, 1346, 4040}
Now we have to determine x and z in terms of y. Using our original equations, we can make the following matrices:
( y 1 ) * ( x ) = ( 2020 )
( 1 y ) ( z ) ( 2021 )
Using these matrices to calculate our system of equation, we get:
x = (2020y - 2021)/(y² - 1) and z = (2021y - 2020)/(y² - 1)
Applying our possible values of y we get only 2 possible solutions:
(x, y, z) = {(2021, 0, 2020); (673, 2, 674)}

[manojsurya]

The factorisation for these Diophantine systems are hard, u did it very well dude😃

[mustafizrahman]

Thanks for sharing. Great explanation.

[siralanturing]

SyberMath:- Shows an interesting System of Equations with *Integer Solutions*
Chinese baby:- My mommy read this out to me as a lullaby

[darkomarkovic]

Interesting and not too difficult problem 👍

[arunavabhar]

X = 1st wave, y = 2nd wave and z = 3rd wave

[ayaannarayan]

Love the videos! Learning a lot! Keep up the good work:)

[aymanehalimi]

Nice 💪
Can you please do more videos about infinite series and their limites, we need them si much at school 🧑‍🎓

[MSJ_]

This was a nice one. However, I have a question on this.... (y-1) (z-x) could be reciprocals of each other as well. y and z being integers, the quantity 1/ (y-x) or 1/(z-x) could be fractions and being the reciprocals, their product would be 1. How will we solve this further is a different matter, but shouldn't this case be considered while solving for (y-1)(z-x) =1 !!