Solve for x,y,z in integer. #x+y+z=1# , #x^3+y^3+z^2=1#?

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Answer 1 · 2016-09-24T08:14:51

#(x,y,z) = (1,-1,1) or (-1,1,1)#

Answer 2 · 2016-09-24T11:17:01

#{y=-3,x=-2,z=6}#
#{y=-2,x=-3,z=6}#
#{y=-2,x=0,z=3}#
#{y=0,x=-2,z=3}#
#{y=0,x=1,z=0}#
#{y=1,x=0,z=0}#

#x+y=1-z#
#x^3+y^3=1-z^2#

Dividing term to term the second equation by the first we have

#(x^3+y^3)/(x+y) = ((1-z)(1+z))/(1-z)# or

#x^2-xy+y^2=1+z#

Adding this equation with the first we have

#x^2-x y+y^2+x+y=2#. Solving for #x# we obtain

#x = 1/2 (-1 + y pm sqrt[3] sqrt[3 - 2 y - y^2])#

Here

#3 - 2 y - y^2 ge 0# so

#-3 le y le 1# but #y in NN# so #y in {-3,-2,-1,0,1}#

Checking we have

#{y=-3,x=-2,z=6}#
#{y=-2,x=-3,z=6}#
#{y=-2,x=0,z=3}#
#{y=0,x=-2,z=3}#
#{y=0,x=1,z=0}#
#{y=1,x=0,z=0}#

for #y = -1# the solutions, are not integer solutions.