30th IMO shortlist 1989/3 solution

30th IMO 1989 shortlist

Problem 3

The polynomial xⁿ + n x^n-1 + a₂x^n-2 + ... + a₀ has n roots whose 16th powers have sum n. Find the roots.

Solution

Answer: all the roots are -1.

Suppose the roots are r₁, r₂, ... , r_n. Note that ∑ r_i = -n. Applying Cauchy to 1, r_i, we have
n² = |∑ r_i|² ≤ (∑ 1²) |∑ r_i²|, so |∑ r_i²| ≥ n.

Applying Cauchy to 1, r_i², we have:
n² ≤ |∑ r_i²|² ≤ ∑ 1² |∑ r_i⁴|, so |∑r_i⁴ ≥ n.

Similarly, applying Cauchy to 1, r_i⁴, we have:
n² ≤ |∑ r_i⁴|² ≤ n |∑ r_i⁸|, so |∑ r_i⁸| ≥ n.

Finally, applying Cauchy to 1, r_i⁸, we have:
n² ≤ |∑ r_i⁸| ≤ n |∑ r_i¹⁶|, so |∑ r_i¹⁶| ≥ n.

But |∑ r_i¹⁶| = n. Hence all the above inequalities are equalities. So, particular, |∑ r_i²| = n. We have equality in Cauchy iff the ratio of the corresponding terms is equal. Hence, considering the first inequality, we must have all r_i equal. Hence all the roots are -1.

30th IMO shortlist 1989