Differences

This shows you the differences between two versions of the page.

--- pow:problem2f22 [2022/09/27 14:01] – mazur
+++ pow:problem2f22 [2022/09/28 02:42] (current) – mazur
@@ Line 1: / Line 1: @@
+<box 85% round orange| Problem 2 (due on Monday, September 26)>
+Find all positive integers $n$ such that $n!$ divides $(2n+1)^{2n}-1$. \\
+(Here $n!=1\cdot 2\cdot \ldots \cdot n$ is the factorial of $n$).
+</box>
+The positive integers in question are $1,2,3,5,6$. Our solution and the two submitted solutions all follow
+the same strategy: show that with a finite and small list of exceptions, the highest power of 2 which
+divides $n!$ is larger than the highest power of 2 which divides $(2n+1)^{2n}-1$, hence $n!$ can not be a divisor of $(2n+1)^{2n}-1$. The small number of exceptions is then handled by hand. For a complete solution and some additional problems and material see the following link {{:pow:2022fproblem2.pdf|Solution}}.