Find all pairs (p,q) of real numbers such that whenever α is a root of x 2 + p x + q = 0 is also root of the equation.

Question

Find all pairs (p,q) of real numbers such that whenever   α is a root of       x    2    +  p  x  +  q  =  0 is also root of the equation.

Eleanor Luna · Accepted Answer

Let   α  ,  β be roots of this polynomial. If   α  =  β, then       α    2    −  2  =  α   α  =  −  1  ,  2, and then   (  p  ,  q  )  =  (  2  ,  1  ) or   (  −  4  ,  4  ), both choices work.Suppose   α  ≠  βIf       α    2    −  2  =  β       β    2    −  2  =  α, then   (      α    2    −  2      )    2    −  2  =  α       α    4    −  4      α    2    −  α  +  2  =  0, same for   β. The polynomial       x    4    −  4      x    2    −  x  +  2 is   (  x  −  2  )  (  x  +  1  )  (      x    2    +  x  −  1  )   α  ,  β  ∈  {  2  ,  −  1  } do not work. So   {  α  ,  β  }  =  {  (  −  1  +      5    )      /    2  ,  (  −  1  −      5    )      /    2  }. It works,   (  p  ,  q  )  =  (  1  ,  −  1  )Finally it could be that       α    2    −  2  =  α which means   α  ∈  {  −  1  ,  2  }, and       β    2    −  2  =  α, so either   α  =  −  1       β    2    −  2  =  −  1       β    2    =  1   β  =  1 since   β  ≠  α   (  p  ,  q  )  =  (  0  ,  −  1  ) or   α  =  2,       β    2    −  2  =  2   β  =  −  2   (  p  ,  q  )  =  (  0  ,  −  4  )Hence the answer:   (  p  ,  q  )  ∈  {  (  −  4  ,  4  )  ,  (  2  ,  1  )  ,  (  1  ,  −  1  )  ,  (  0  ,  −  1  )  ,  (  0  ,  −  4  )  }

Find all pairs (p,q) of real numbers such that whenever α is a root of

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