 Solution

ID: past-exam-of-the-analytic-number-theory-and-elliptic-curves-course-of-the-university-of-oxford/2007/b/5/ii/solution

\sum_{j \geq 0} p^{2 j}

converges to

1/ (1 - p^{2})

. Consequently

\frac{1 + 2 p}{p - p ^{3}} = (p^{- 1} + 2) \sum_{j \geq 0} p^{2 j} = p^{- 1} + 2 + p + 2 p^{2} + p^{3} + 2 p^{4} + \dots .

Since

p \neq = 2

, the digits one and two are both between zero and

p - 1

, so no carrying is required. These are its p-adic digits, with the unique negative-power digit at exponent minus one.

 New to topics? Read the docs here!