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.

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step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)

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step-by-step explanation:

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