Solve for X , Y , Z where : X ² = Y + a Y ² = Z + a Z ² = X + a

Question

Solve for   X  ,  Y  ,  Z where :       X      ²    =  Y  +  a      Y      ²    =  Z  +  a      Z      ²    =  X  +  a

fugprurgeil · Accepted Answer

You can turn it into a single polynomial

X = Z^{2} - a Y = (Z^{2} - a)^{2} - a ((Z^{2} - a)^{2} - a)^{2} = Z + a

This is a degree-8 polynomial in

Z

, so there isn't a formula for its solution, but software can solve it.
Two of the solutions are the solutions to

Z^{2} = Z + a

, which can be solved with the quadratic formula. The other six solutions form triples

Z, Z^{2} - a, (Z^{2} - a)^{2} - a

According to Galois theory, the symmetry that the triples have makes it simpler. They obey cubics whose coefficients are quadratic in

a

. If

b

and

c

are the roots of

x^{2} + x + 2 = a

, then I think the original octic factors in this way:

(Z^{2} - Z - a) (Z^{3} + (1 + b) Z^{2} + (b - a) Z - 1 - a b) . (Z^{3} + (1 + c) Z^{2} + (c - a) Z - 1 - a c)

So you can get exact solutions as a function of

a

, if you know how to solve cubics.

Solve for X , Y , Z where : X ² </msup> = Y +