a) Well, any real number has one of three signs: 0, + or -.
x, y, and z cannot be of three different signs, because (x + y)^5 would be negative, but z is positive.
Similarly, two of them cannot be of the same sign that differs from the 3rd, because the (sum of those two)^5 has the same sign as those two.
So, x, y and z are necessarily of the same sign.
b) Clearly, (0,0,0) is one solution
c) Assume that there is another solution, where they are all positive,
(y+z)^5 = x <= z = (x+y)^5
But (y+z)^5 <= (x+y)^5 only if z = x, so all three are equal.
Then (2x)^5 = x
32x^5 = x
x^4 = 1/32
x = (1/32)^(1/4)
Thus, there is only one all-positive solution,
((1/32)^(1/4),(1/32)^(1/4),(1/32)^(1/4))
d) Similarly, there is only one all-negative solution:
(-(1/32)^(1/4),-(1/32)^(1/4),-(1/32)^(1/4))
So the only solutions are
(0,0,0)
((1/32)^(1/4),(1/32)^(1/4),(1/32)^(1/4))
(-(1/32)^(1/4),-(1/32)^(1/4),-(1/32)^(1/4))
Edited on October 27, 2013, 8:56 am
Got them all! (spoiler)
