### 12th Putnam 1952

Problem A3

Let the roots of the cubic p(x) ≡ x3 + ax2 + bx + c be α, β, γ. Find all (a, b, c) so that p(α2) = p(β2) = p(γ2) = 0.

Solution

Answer: (0, 0, 0) roots 0, 0, 0
(-1, 0, 0) roots 0, 0, 1
(-2, 1, 0) roots 0, 1, 1
(0, -1, 0) roots 0, 1, -1
(1, 1, 0) roots 0, ω, ω2, where ω = e2pi/3
(-3, 3, -1) roots 1, 1, 1
(-1, -1, 1) roots 1, 1, -1
(1, -1, -1) roots 1, -1, -1
(0, 0, -1), roots 1, ω, ω2
(i, 1, -i), roots 1, -1, -i
(-i, 1, i), roots 1, -1, i
(1-ω, 1-ω, -ω, roots ω, ω, ω2
(1-ω2, 1-ω2, -ω2), roots ω, ω2, ω2
(1+ω, 1+ω, ω), roots ω, -ω, ω2
(1+ω2, 1+ω2, ω2), roots ω, ω2, -ω2
((1-i√7)/2, (-1-i√7)/2, -1), roots k, k2, k4, where k = e2iπ/7
((1+i√7)/2, (-1+i√7)/2, -1), roots k3, k5, k6.

p(x) = 0 has only three roots, so the squares must all be found in { α, β, γ }. Let α be the root with the largest modulus, then |α|2 > |α| unless |α| ≤ 1. Let γ be the root with the smallest non-zero modulus, then |γ|2 < |β| unless |γ| ≥ 1. Hence all non-zero roots have modulus 1.

All roots zero is possible and gives (a, b, c) = (0, 0, 0).

Suppose β = γ = 0 and α is non-zero. Then α2 = α, so α = 1. This gives (a, b, c) = (-1, 0, 0).

Suppose γ = 0, and α, b are non-zero. If α2 = α, then α = 1. Then β2 = 1 or β, so β = 1 or -1. This gives (a, b, c) = (-2, 1, 0) or (0, -1, 0). If α2 is not α (and β2 not β), then we must have α4 = α, so α and β are ω and ω2, where ω = e2iπ/3. This gives (a, b, c) = (1, 1, 1).

The remaining possibility is that all roots are non-zero (and have modulus 1). There might be 3, 2, 1 or 0 roots equal to 1. The first case gives (-3, 3, -1). In the second case the third root must satisfy γ2 = 1 or γ and hence must be -1, giving (a, b, c) = (-1, -1, 1).

We consider the third case (just one root equal to 1). If both the roots not equal to 1 have square 1, then they must both be -1 and (a, b, c) = (1, -1, -1). If one has square 1 and one not. Then they must be -1 and ±i. This gives (a, b, c) = (i, 1, i) or (-i, 1, -i). If neither have square 1, then we must have β2 = γ, γ2 = β, hence β3 = 1 and β and γ are ω and ω2.

The final case is that no roots are 1. If we have α2 = β and β2 = α, then α and β are ω and ω2 and γ2 = ω or ω2, so γ = ±ω or ±ω2. So the roots are ω, ω, ω2 or ω, ω2, ω2 or ω, -ω, ω2 or ω, ω2, -ω2. This gives (a, b, c) = (1-ω, 1-ω, ω) or (1-ω2, 1-ω2, ω2) or (1+ω, 1+ω, ω) or (1+ω2, 1+ω2, ω2).

The final possibility is α2 = β, β2 = γ, γ2 = α. Hence α7 = 1. If we put k = ei2π/7, then the roots could be k, k2, k4 or k6, k5, k3. This corresponds to (a, b, c) = ((1-i√7)/2, (-1-i√7)/2, -1), ((1+i√7)/2, (-1+i√7)/2, -1).

Comment. In some ways this is a nightmare question to meet in an exam. It is not particularly difficult, but it is almost impossible to do quickly and there is much scope for minor mistakes.