 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Symmetry helps! (Posted on 2016-10-26) Solve the system of equations below, listing all un-ordered triplets of integers:

x^2 −(y+z+yz)x+ (y+z)yz = 0
y^2 −(z+x+zx)y+ (z+x)zx = 0
z^2 −(x+y+xy)z+ (x+y)xy = 0

 See The Solution Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) my short solution | Comment 4 of 7 | Let p=y+z and q=y+z

Then the 1st equation becomes  x^2-(p+q)*x+p*q=0

equivalent to  (x-p)*(x-q)=0

which implies  x1=p  x2=q  i.e.  x=y+z  or x=y*z

By symmetry two other equations force  y=x+z  or y=x*z

and z=x+y or  z=x*y.

Clearly, out of 8 possible combinations of "binary triplets" only

(0,0,0) and (1,1,1) qualify.

 Posted by Ady TZIDON on 2016-10-27 04:20:49 Please log in:
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