Show that if n is a composite integer not equal to 4, then (n-1)! is congruent to 0 (mod n).
Please help... Thanks.
if n = a^2, and n > 4, then a > 2. so n = a^2 > 2a > a > 2 , in which case n-1 ≥ 2a > a > 1, so both a and 2a are factors of (n-1)! since a^2 divides 2a^2, a^2 clearly divides (n-1)!
for n = 4 (a = 2), 2a is "too big" to divide 3!, 4 does not divide 6.
for n = 9 (a = 3), 3 and 6 (a and 2a) are the only factors of 8! = (2)(3)(4)(5)(6)(7)(8) divisible by 3, a is "just small enough".