Finding Positive Integer Solutions to an Equation (x+y+z=xyz)

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#NumberTheoryProblems #DiophantineEquations
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[-basicmaths]

Here x, y, z are not equal so i take x<y<z. You can also write other options.

[MathElite]

Nice number theory question, I'm trying to learn more about the subject

[josephsilver]

You always make it seem so easy! Good video.

[deepjyoti]

Thnx, u add something new to my knowledge

[beagleboysinc]

I've always (almost always) known that 1, 2 and 3 satisfied that equation. Nice to see a proof that those are the only numbers that qualify.

[razor-xnve]

Here is a short cut
Divide both sides by xyz
1/xy +1/yz +1/zx =1
Now sum of three terms is 1
By contradiction atleast one must be greater than 1/3
Let 1/xy>1/3
So xy<3
Rest is a piece of 🍰

[-basicmaths]

Here I have a, b as
y=ax, z=bx
x<y<z then b>a>1

[srijanbhowmick]

Here are 3 suggestions for your channel :----
a) Solve a hard geometry puzzle every month (a personal request from me bcoz I loved them very very much, also one of the reasons I decided to stick with your channel )

b) Solve a viewer suggested problem every 15 days if possible

c) Introduce new type of problems in your channel (yeah it's getting a tad bit repetitive)

Also hey I don't comment every now and then nowadays but I still watch every single one of ur videos and bcoz of u I have been able to solve
all of your problems over the last 2 months so I wanna thank you once more for ur contributions to the math community . Also plz tell ur age and nationality 😁

[math]

great job Syber, thanks for sharing bro

[marioszaros]

Another inequality that can be applied is at (x+y)/(xy-1)=z>/1 =>
2>/ (z-1)(y-1) and from there the maximum value of z, y is 3 and also x since its symmetrical

[aouab]

i cant believe this. i was thinking about this question for the past 2 days and kept looking for solutions. and guess what? i get a video for it omg

[rnite]

Imma see if I can solve this.
Or_0 also keep up the great content Syber.

[-basicmaths]

x+y+z=xyz
x, y, z are positive integers, then
We have y=ax, z=bx
x+ax+bx=abx^3
x(1+a+b)=abx^3
(1+a+b)=abx^2
x^2=(1+a+b)/ab
Now RHS is less than 3
x^2=2, x^2=1, as x is positive integer
So x=1, y=a, z=b
1+a+b=ab
1+a=ab -b,
1+a=b(a-1)
b=1+a, a-1=1
a=2, b=3
(x, y, z)=(1, 2, 3)

[-basicmaths]

In this equation x, y, z are not equal
If they are equal then
3x=x^3
x^2=3 so x isn't integer.
If two of them are same then
2x+z=x^2z
2x=x^2z-z=z(x^2-1)
z=2x/(x^2-1). from this We have z
isn't integer.so x, y, z are not equal.

[yasharthyash]

I used trigo for solving this using tangent in place of x, y,z.. i also got the solution by getting equations x+y=z... and from that i get xy=2, and as positive asked then apply simple no. Theory

[aliasgharheidaritabar]

Thank you for your nice and instructive videos.
I learn alot in your channel.

[abgd]

THANK YOU SO MUCH, with your help i solved my first problem in number theory ♥️♥️♥️♥️

[michaelempeigne]

when @sybermath says that x, y, and z are interchangeable; he in essence means that there are symmetric solutions.

[nalapurraghavendrarao]

Set of numbers which satisfy the condition that their sum is equal to their product are called perfect numbers.will you please enlighten us about other sets of numbers.

[tonyhaddad]

Nice problem but quite easy (1, 2, 3 ) with permutation ( proof by inequality) !!!