From A Squared Divisor we know for all natural numbers n>=2 that n^(n-1)-1 is divisible by (n-1)^2.

Prove that n=2 and n=3 are the only natural numbers such that n^(n-1)-1 is divisible by (n-1)^3.

To solve the given problem, we need to prove that $n=2$ and $n=3$ are the only natural numbers such that:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mi>n</mi><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msup><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">n^{(n-1)} - 1</annotation></semantics></math>n(n−1)−1

is divisible by <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><msup><mo stretchy="false">)</mo><mn>3</mn></msup></mrow><annotation encoding="application/x-tex">(n-1)^3</annotation></semantics></math>(n−1)3.

Edited by uno online 1 day ago

Posted by Eddie Nelson on 2025-03-28 02:39:56