Prove: for any natural number n, either n is prime, n is a perfect square, or n|(n-1)!

Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n, q < n , p ≠ q and pq = n.

Now, note that (n - 1)! = (n - 1)(n - 2)(n - 3)...(3)(2)(1).

Since p < n and q < n, p = (n - k) and q = (n - l) where k ≠ l and k, l ∈ {∈1, 2, 3, ... , n - 1}. But we know that (n - k)(n - l) | (n - 1)!

Hence, pq | (n - 1)! and therefore n | (n - 1)!

Hence, any natural number n is either prime, a complete square or divides (n - 1)!, as required.

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225

step by step explanation: when u removing the parentheses and combined them that would be : x+3y-x-y=2y