Here's the question you clicked on:
Ishaan94
Find all real triples \((x,y, z)\) that satisfy \(x^4+y^4+z^4−4xyz = −1\).
It is just a surface in a 3 dimensional space.
I've seen a method to solve problems like this easily, but I can't remember it right now. I'll try and look it up later to try and find more information.
|dw:1341442265927:dw|
|dw:1341442357328:dw|
|dw:1341442492121:dw|
So I just found out min(f)=-1 so as I told in above that is a surface in a space.
hi @Ishaan94 Using Completing the Square \[x^4+y^4+z^4-4xyz+1=0\\x^4-2x^2y^2+y^4+2x^2y^2+z^4+2z^2-2z^2-4xyz+1=0\\x^4-2x^2y^2+y^4+2(x^2y^2-2xyz+z^2)+z^4-2z^2+1=0\\(x^2-y^2)^2+2(xy-z)^2+(z^2-1)^2=0\] so \[x^2-y^2=xy-z=z^2-1=0\]
only triples that satisfies the equation \[(x,y,z)=(1,1,1),(−1,−1,1),(1,−1,−1),(−1,1,−1)\]
Thanks @mukushla if it's not too much to ask did you take training or something for the olympiads?
welcome my friend no im a chemical engineering student and just love math
do u wanna get ready for Math Olympiad or something?
no i am already past my high school but i love solving math problems.