Find all pears (x,y,z) (real number) such that
x^4 + y^4 + z^4 = 1

Question

Find all pears (x,y,z) (real number) such that
x^4 + y^4 + z^4 = 1

hero · Answer

Okay, this should work

Let 

x = $\sqrt[4]{\frac{1}{7}}$
y = $\sqrt[4]{\frac{2}{7}}$
z = $\sqrt[4]{\frac{4}{7}}$

hero · Answer

Good luck finding the rest of them

hero · Answer

Me personally, I think it's infinite because you can always find three fractions that add to get 1

anonymous · Answer

If you want Integer solutions, there are only these :
$$x = \pm1; y = 0; z = 0$$
$$x = 0; y = \pm1; z = 0$$
$$x = 0; y = 0; z = \pm1$$
Observe that a number raised to the 4th power is always positive. Therefore only one variable can have the value 1, and the rest must be 0

hero · Answer

He never mentioned "integers" in the question, but okay.

hero · Answer

"all pairs of real numbers" doesn't imply "integers only"

anonymous · Answer

Yes, I know. I'm just giving additional info :)

hero · Answer

By the way, those numbers I posted can also be negative

hero · Answer

@RadEn, do you have anything to add or not. I'm about to hit the hay

raden · Answer

sorry, my conections very very slow, lost lost lost ohh :(
Ok thank u..
yea, i got.. 
so, i should find the combin the sum of 0<=x^4+y^4<=1, and z will be take them

hero · Answer

The numbers can be less than zero, however, they will still be fractions

anonymous · Answer

to be precise :
-1 <= x, y, z <= 1

raden · Answer

oh yea, negative number power even number must be positive... ok thank u

anonymous · Answer

one wonders how $(x,y,z)$ constitutes a pair

hero · Answer

family*