Polynomial

[R.D. Silverman]

Originally Posted by kubus

No way. Constant coeff in %9 is -64, so there must be k⁶=-64 if we want a sextic in (z + k/z).

??

We need k^6 + c1 k^5 + c2 k^4 + ... = -64

[kubus]

Originally Posted by R.D. Silverman

??

We need k^6 + c1 k^5 + c2 k^4 + ... = -64

No.
There must be satisfied
z⁶ ( (z + k/z)⁶ + c₁ (z + k/z)⁵ + c₂ (z + 2/z)⁴ + ... ) = z¹² + ... - 448z - 64
The only constant term on the left side is k⁶. The others are divisible by z.

[R.D. Silverman]

Originally Posted by kubus

No.
There must be satisfied
z⁶ ( (z + k/z)⁶ + c₁ (z + k/z)⁵ + c₂ (z + 2/z)⁴ + ... ) = z¹² + ... - 448z - 64
The only constant term on the left side is k⁶. The others are divisible by z.

Yup.