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--- pow:problem2 [2020/03/02 22:34] – mazur
+++ pow:problem2 [2020/03/02 22:55] (current) – mazur
@@ Line 1: / Line 1: @@
+<box 80% round orange|Problem 2 (due Monday, March 2)>
+Recall that the smallest integer greater or equal than a given real number $x$ is denoted by $\left \lceil
+x \right\rceil $ and called the $ceiling$ of $x$. Let $p$ be a prime number and $1\leq a<p$ an integer.
+Prove that the number
+\[ \left \lceil  \left(a^{p-1}-1 \right)^{\frac{p}{p-1}} \right\rceil \]
+is  divisible by $p$. What can you say when $a=p$?
+</box>
+This problem was solved by only one participant: Yuqiao Huang. The submitted solution does not discuss the case when $a=p$. For $1\leq a\leq p-1$, the solution claims correctly that
+$\left \lceil  \left(a^{p-1}-1 \right)^{\frac{p}{p-1}} \right\rceil= a^p-a$, hence the result follows
+from Fermat's Little Theorem. In order to justify the claim, the solver proves that
+\[ a^p-a-1< \left(a^{p-1}-1 \right)^{\frac{p}{p-1}}\leq a^p-a.\]
+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 {{:pow:2020sproblem2.pdf|Solution}}