We can assume that p ≠ 2 , 3 because 0 is a root of f ( x ) in F 2 and F 3 . Let Q be the "squares subgroup" (?) of index 2 in F p ∗ . If 2 and 3 are not in Q (this means that ( x 2 − 2 ) ( x 2 − 3 ) doesn't have roots in F p ), then 2 ⋅ 3 ∈ Q (?), and then x 2 − 6 has a root in F p

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We can assume that   p  ≠  2  ,  3 because 0 is a root of   f  (  x  ) in             F        2   and             F        3  . Let Q be the &quot;squares subgroup&quot; (?) of index 2 in             F        p    ∗  . If 2 and 3 are not in Q (this means that   (      x    2    −  2  )  (      x    2    −  3  ) doesn&#039;t have roots in             F        p  ), then   2  ⋅  3  ∈  Q (?), and then       x    2    −  6 has a root in             F        p

Karla Hull · Accepted Answer

By the squares subgroup, they mean the subgroup  Q  =  {  a  ∈            F        p    ∗    ∣  a  =      b    2     for some   b  ∈            F        p    ∗    }  .In other words, Q is the image of the homomorphism   φ  :            F        p    ∗    →            F        p    ∗   given as   x  ↦      x    2  In the below, I am also going to assume that   p  ∉  {  2  ,  3  }Since   ker  ⁡  φ  =  {  1  ,  −  1  } and   −  1  ≠  1 (why?), we see that G has index 2 in             F        p    ∗  Now, if   2  ,  3  ,  6  ∉  Q, then this means that the cosets   2  Q  ,  3  Q  ,  6  Q are all not equal to Q. Since Q has index 2 in             F        p  , this means that   2  Q  =  3  Q  =  6  QThus, one of   2  ,  3  ,  6 must be in Q. In turn, the given polynomial has a root in             F        p   (in fact, in             F        p    ∗  , if   p  ≠  2  ,  3)

We can assume that p ≠ 2 , 3 because 0 is a root of f ( x ) i

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