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pow:problem2

Problem 2 (due Monday, March 2)

Recall that the smallest integer greater or equal than a given real number x is denoted by x and called the ceiling of x. Let p be a prime number and 1a<p an integer. Prove that the number

(ap11)pp1 is divisible by p. What can you say when a=p?

This problem was solved by only one participant: Yuqiao Huang. The submitted solution does not discuss the case when a=p. For 1ap1, the solution claims correctly that (ap11)pp1=apa, hence the result follows from Fermat's Little Theorem. In order to justify the claim, the solver proves that apa1<(ap11)pp1apa. The proof of the right hand side inequality is fairly simple; the submitted proof of the left hand side inequality is rather long and complicated, and we will not reproduce it here. To see a detailed solution click the following link Solution

pow/problem2.txt · Last modified: 2020/03/02 17:55 by mazur