All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Only one, one, one (Posted on 2007-05-30)
It's obvious that x=y=z=1 is an all-positive, all-real, solution of

x + y² + z³ = 3
y + z² + x³ = 3
z + x² + y³ = 3

Can you prove that it is the only such solution?

 See The Solution Submitted by Federico Kereki Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 5
Let's assume there is a solution which can be written as x = 1+d_x, y=1+d_y, z=1+d_z. The first equation, after some tinkering, then becomes d_x + 2 d_y + d_y^2 + 3d_z + 3 d_z^2 + d_z^3 = 0. The other equations give a similar result. Summing all three of these results gives:
d_x^3 + 4d_x^2+6d_x
+ d_y^3 + 4d_y^2+6d_y
+d_z^3 + 4d_z^2 + 6d_z = 0

The first part (the part with d_x) can be factored as d_x(d_x^2+4d_x+6). The only zero of this is d_x=0. The same goes for the other parts. The quadratic part will always be strictly positive, so the only solution of the equations is d_x=d_y=d_z=0, which results in x=y=z=1.

 Posted by Robby Goetschalckx on 2007-05-31 03:16:07

 Search: Search body:
Forums (0)