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.
You have five coins, apparently alike, but actually of different weights. You also have a two arm scale.
Can you manage to sort the coins in ascending order, using the scale only seven times?
Bonus question: can it be done in fewer weighings?
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Bonus: 
