### 48th Putnam 1987

Problem B3

F is a field in which 1 + 1 ≠ 0. Define Pα = ( (α2 - 1)/(α2 + 1), 2α/(α2 + 1) ). Let A = { (β, γ) : β, γ ∈ F, and β2 + γ2 = 1}, and let B= { (1, 0) } ∪ { Pα: α ∈ F, and α2 ≠ -1}. Prove that A = B.

Solution

Fairly easy

Obviously B ⊆ A. The problem is to find some expression for α in terms of β and γ.

We start from β = (α2 - 1)/(α2 + 1), and get α2 = (1 + β) / (1 - β). (*)

We cannot take square roots. So it may not be obvious what to do next. But we do have one square root available, because we know that γ2 = 1 - β2. It is now obvious that we multiply top and bottom of (*) by (1 + β). That gives α = (1 + β)/γ.

So we can now start again to show that A ⊆ B. Given (β, 0) ∈ A. Then β2 - 1 = 0, so (β + 1)(β - 1) = 0, so β = ±1. (1, 0) is specifically included in B, and (-1, 0) = P0. Now for (β, γ) ∈ A with γ ≠ 0, we also have β ≠ 1. Let α = (1 + β)/γ. Then α2 = (1 + β)22 = (1 + β)2/( (1 + β)(1 - β) ) = (1 + β)/(1 - β). Hence α2 + 1 = 2/(1 - β), and α2 - 1 = 2β/(1 - β), so Pα = (β, γ).