Giuga Numbers are Squarefree #
If n is a Giuga number, then n is squarefree.
Proof: Suppose p² ∣ n for some prime p ∣ n. Then p ∣ (n/p).
But the Giuga condition gives p ∣ (n/p - 1). So p ∣ (n/p) - (n/p - 1) = 1,
contradicting p ≥ 2.
If n is a Giuga number, then n is squarefree.
Proof: Suppose p² ∣ n for some prime p ∣ n. Then p ∣ (n/p).
But the Giuga condition gives p ∣ (n/p - 1). So p ∣ (n/p) - (n/p - 1) = 1,
contradicting p ≥ 2.