MathLabs

Problem 3

Let p≥5p\ge5 be a prime. Let rr be the number of ways of placing pp identical checkers on a p×pp\times p checkerboard so that not all checkers are in the same row (they may all be in the same column). Show that rr is divisible by p5p^5.
Step 5 of 5: Conclude divisibility
p5∣rp^5\mid r
Detailed analysis

The product congruence says the parenthesized factor in r=(p2p)−pr=\binom{p^2}{p}-p is divisible by p4p^4. The leading factor pp then gives p5∣rp^5\mid r, as required.