Discrete Algebraic Methods

The derivative of t is t(X) = 12(X3 + AX + B). The polynomial t' has no multiple roots because X3 + AX + B has three distinct roots. Therefore, t has no triple roots. Suppose that t(X) has two different double roots c ≠ d. Then, t(X) = 3(X−c)2(X−d)2 = 3(X2 − 2cX + c2)(X2 − 2dX + d2). Comparing coefficients at X3 yields 0 = −6(c + d). Then, c = −d and therefore t(X) = 3(X2 − c2)2 = 3X4 − 6c2X2 + 3c4. This implies B = 0 as well as A = −c2 and A2 = −3c4. We obtain A = B = 0, which contradicts 4A3 + 27B2 ≠ 0. Therefore, t has at least two simple roots c1 ≠ c2. We further have for i ∈ {1, 2}. Thus for suitable di ≠ 0 we can find four different points ...