 Solution

ID: past-exam-of-the-number-theory-course-of-the-university-of-oxford/2009/a/1/d/solution

Y = X + 1

. The polynomial becomes

Y^{4} + Y + 1

, and translating all roots leaves their pairwise differences unchanged. Part (b) gives

Δ_{f} = 256 - 27 = 229.

It is nonzero and lies strictly between

1 5^{2} = 225

and

1 6^{2} = 256

. An integer which is a rational square is an integer square, so this is not a rational square. The roots are distinct because the discriminant is nonzero. By (c),

G_{f} \neq \subseteq A_{4}

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