4:50 if z=1, then xy=1 gives x=y= 1 OR x=y=-1. ie a fourth solution. This corresponds with the fact that the equation is symmetric in x,y,z and so the solution must be symmetric. ie the solution set is x=y=z=1 and all permutations of (x,y,z)=(1,1,-1)
I used AM>=GM formula: (x^4 + y^4 + z^4 + 1)/4 = 4throot(x^4 * y^4 * z^4 * 1). It works, as x^4, y^4 and z^4 are always nonnegative. After a few steps, you get x^4 + y^4 + z^4 >= 4xyz - 1. So, the relation becomes an equation, if x^4 = y^4 =z^4. It is still a little tricky, as you have to be careful with the signs. It's not the same as x=y=z apparently, as it only gives the solution (1,1,1)... I'm not sure how you could find the rest of solutions in an universal way from there. Or you could just try out all 3 possibilities with +/- 1 and see, what happens.
I wonder how at 4:48 you could have missed the solution (−1, −1, 1) of your equation in x, y, z when at the end you show the result from WA which clearly finds _four_ integer solutions.
The equation is symmteric over x,y,z So if there is a solution {1,-1,-1} there should be the solutions {-1,1,-1} and {-1,-1,1}. You did miss one. And definitely the complex solutions should be considered as well.
I found two videos where you already did both problems from this video, but UA-cam won't let me post the links because if I do my message gets removed immediately. But anyway, here are the video IDs: ogfNGLyL7NA pFUVGjMUHMY
Thanks for the heads up! 😍 One of them is a generalization or a version with four variables. The other one is the same problem done slightly differently, maybe! 😜 Here's the links: ua-cam.com/video/ogfNGLyL7NA/v-deo.html ua-cam.com/video/pFUVGjMUHMY/v-deo.html
Proving a quartic inequality (Sum of Fourth Powers): ua-cam.com/video/nCburjt1UgU/v-deo.html
4:50 if z=1, then xy=1 gives x=y= 1 OR x=y=-1. ie a fourth solution.
This corresponds with the fact that the equation is symmetric in x,y,z and so the solution must be symmetric.
ie the solution set is x=y=z=1 and all permutations of (x,y,z)=(1,1,-1)
I used AM>=GM formula: (x^4 + y^4 + z^4 + 1)/4 = 4throot(x^4 * y^4 * z^4 * 1). It works, as x^4, y^4 and z^4 are always nonnegative. After a few steps, you get x^4 + y^4 + z^4 >= 4xyz - 1. So, the relation becomes an equation, if x^4 = y^4 =z^4. It is still a little tricky, as you have to be careful with the signs. It's not the same as x=y=z apparently, as it only gives the solution (1,1,1)... I'm not sure how you could find the rest of solutions in an universal way from there. Or you could just try out all 3 possibilities with +/- 1 and see, what happens.
Good to see you! ☺️
@@SyberMath Hi! I often follow your content, but only comment on problems not too obvious and not too difficult for me. ;-)
4:50 you missed (-1, -1, 1)
😄😄
*Yeah, a quartic is supposed to have four solutions*
I wonder how at 4:48 you could have missed the solution (−1, −1, 1) of your equation in x, y, z when at the end you show the result from WA which clearly finds _four_ integer solutions.
😮😄 that's me!
arithmetic mean >= geometric mean yields (1, 1, 1)
other solutions are (1, -1, -1) etc.
The equation is symmteric over x,y,z
So if there is a solution {1,-1,-1} there should be the solutions {-1,1,-1} and {-1,-1,1}.
You did miss one.
And definitely the complex solutions should be considered as well.
Complex solutions examples:
{-1,i,i},{1,i,-i}
because
i⁴=1
-1•i•i=1
1•i•(-i)=1
😄😄
Bro why are some people hating so hard in the comments when he accidentally missed a solution
Me seeing the problem, be like:
"it's AM GM, it's AM GM"
Giving this kind of question is Cheating, because the question maker already knows the answer.
ehehehehehe 😄😜
I found two videos where you already did both problems from this video, but UA-cam won't let me post the links because if I do my message gets removed immediately. But anyway, here are the video IDs:
ogfNGLyL7NA
pFUVGjMUHMY
Thanks for the heads up! 😍
One of them is a generalization or a version with four variables. The other one is the same problem done slightly differently, maybe! 😜
Here's the links:
ua-cam.com/video/ogfNGLyL7NA/v-deo.html
ua-cam.com/video/pFUVGjMUHMY/v-deo.html