Let b , c ∈ Z and let n ∈ N , n ≥ 2. Let f ( x ) = x n − b x + c. Prove that disc ( f ( x ) ) = n n c n − 1 − ( n − 1 ) n − 1 b n .

Question

Let   b  ,  c  ∈      Z   and let   n  ∈      N  ,   n  ≥  2. Let   f  (  x  )  =      x          n        −  b  x  +  c. Prove that      disc    (  f  (  x  )  )  =      n          n            c          n      −      1        −  (  n  −  1      )          n      −      1            b          n        .

Lillianna Salazar · Accepted Answer

Here is a brute force approach:      f    ′    (  x  )  =  n      x          n      −      1        −  b, and we want to compute       ∏    i        f    ′    (      α    i    ). We do this by looking for the minimal polynomial with roots       α    i          n      −      1      .Note that                              x          n                −        b        x        +        c        =        0                    ⇔        x        (                  x                      n            −            1                          −        b        )        =        −        c                                  ⇔                  x                      n            −            1                          (                  x                      n            −            1                          −        b                  )                      n            −            1                          =        (        −        1                  )                      n            −            1                                    c                      n            −            1                              let       y    i    =      α    i          n      −      1      , and       z    i    =      f    ′    (      α    i    )  =  n      y    i    −  b. We have found the minimal polynomial for       y    i  :  y  (  y  −  b      )          n      −      1        =  (  −  1      )          n      −      1            c          n      −      1      and we want the product       ∏    i        z    i  . Consider the change of variable   z  =  n  y  −  b, i.e.   y  =            z      +      b        n  . Substitute into the above equation, we get            z      +      b        n              (                        z          −          (          n          −          1          )          b                n            )              n      −      1        =  (  −  1      )          n      −      1            c          n      −      1          ⇔  (  z  +  b  )  (  z  −  (  n  −  1  )  b      )          n      −      1        −  (  −  1      )          n      −      1            n    n        c          n      −      1        =  0The constant term of this polynomial in z is   (  −  1      )    n        ∏    i        z    i  , therefore      ∏    i        z    i    =  (  −  1      )    n        (    (    −    1          )              n        −        1                    b      n        (    n    −    1          )              n        −        1              −    (    −    1          )              n        −        1                    n      n              c              n        −        1              )    =      n    n        c          n      −      1        −  (  n  −  1      )          n      −      1            b          n

Let b,c in ZZ and let n in NN, n>=2. Let f(x)=x_n−b_x+c. Prove that disc(f(x))=n^nc^(n−1)−(n−1)^(n−1)b_n.

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