Past exam of the number theory course of the University of Oxford / 2007 / A / 1

By Vieta's formulas, the elementary symmetric polynomials in the roots giveFor each root, ${\alpha }_j^3+c_1{\alpha }_j^2+c_2{\alpha }_j+c_3=0$. Multiplying by ${\alpha }_j^k$ and summing gives the power sum recurrenceIts initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are polynomials with integer coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so induction proves the same for every $s_k$. This is the cubic case of Newton's identities.

For $c_1=0$, put ${\alpha }_1=u$, ${\alpha }_2=v$, ${\alpha }_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences givesThis is the discriminant of a depressed cubic in depressed form, and the displayed identity also proves its expression in integer-coefficient symmetric polynomials.

Scaling every root scales its elementary symmetric polynomials by the corresponding powers of $d$. Therefore the monic polynomial isThis remains valid for $d=0$, when all three roots are zero; deriving it by symmetric coefficients avoids dividing by $d$.

Reciprocal roots require $c_3\ne 0$, equivalently every original root is nonzero. Their monic polynomial is the normalized reciprocal polynomialIf $c_3=0$, at least one reciprocal in the source is undefined, so that part has no polynomial interpretation without the nonzero-root qualification.

The sum of squared roots is $c_1^2-2c_2$. The sum of their pairwise products isTheir product is $c_3^2$. Thus the required monic polynomial is