First, solve the system in terms of k
x + y*z = k
y + x*z = k
z + x*y = k
Take the difference of the first two equations
(x + y*z) - (y + x*z) = k - k
(x-y) + z*(y-x) = 0
(x-y)*(1-z)=0
x=y OR z=1
case x=y OR case z=1
-----------------------------------------------||--------------------------
Substitute x=y into the original system ||Substitute z=1 in the
x + x*z = k AND z + x*x = k ||original system
||
Take the difference of these equations || x + y = k
(x-z) + x*(z-x) = 0 || x * y = k-1
x=z OR x=1 ||
subcase x=z OR subcase x=1 ||Then (x,y,z) = (k-1,1,1)
-----------------------|-----------------------||or (x,y,z) = (1,k-1,1)
x,y,z are all equal |x and y equal 1 ||
solve x^2+x = k |then (x,y,z)=(1,1,k-1) ||
x=(-1+/-sqrt[4k+1])/2 | ||
There are five possible solutions total:
(x,y,z)=( 1, 1,k-1)
(x,y,z)=( 1,k-1, 1)
(x,y,z)=(k-1, 1, 1)
x=y=z = (-1 + sqrt[4k+1])/2
x=y=z = (-1 - sqrt[4k+1])/2
If k=2, there are two solutions (1,1,1) or (-2,-2,-2).
If k>-1/4 and k!=2, there are five solutions.
If k=-1/4, there are four solutions
If k<-1/4, there are three real solutions (and two complex solutions).
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