There exists a polynomial P of degree 5 with the following property: If z is a complex number such that z5+2004z=1, then P(z2)=0. Calculate P(−1)P(1).
Solution — click to reveal
Let r1,r2,r3,r4,r5 be the roots of Q(z)=z5+2004z−1. Then Q(z)=(z−r1)(z−r2)(z−r3)(z−r4)(z−r5)and P(z)=c(z−r12)(z−r22)(z−r32)(z−r42)(z−r52)for some constant c.