Can (n-1) be Divisible by n for Any Natural Number n ≥ 5 and Not Prime?

[xax]

I need to prove this for any n natural, n>= 5, n not prime.

 

[dodo]

Think a bit about the prime factors of n... are they smaller than n? Then think of what (n-1)! means.

 

[HallsofIvy]

Dodo's hint is excellent- but the crucial point is whether the factors of n are less than n-1, not jus n itself!

 

[mathman]

If n is not prime, then n=j*k, where j and k are >1, therefore both <n. As a result both are factors in (n-1)!, so n divides (n-1)!.

 

[rodigee]

Almost there, one last thing to consider is the squares of prime numbers. What happens when n=9=3*3 for example? This is where the n>=5 comes in.

 

[xax]

You all have good points and I thought of all of them and the reason I've posted this is because of the square numbers as rodigee said. Well I think if n = k*k then k is one of the numbers between 1 and n-1. since n=k*k then k divides (n-k) and this is smaller the n-1 which means n divides (n-1)!.

 

[rodigee]

Sorry, xax but I don't really understand your idea heh. This is how I see it:

$$n > 4$$ implies that

$$\sqrt{n} > 2$$ so we also have that $$\sqrt{n}\sqrt{n} > 2\sqrt{n}$$ or simply $$n > 2\sqrt{n}$$, meaning that $$2\sqrt{n}$$ and $$\sqrt{n}$$ both show up in $$(n-1)!$$ so we're good to go.

 

[xax]

You make perfect sense rodigee and this is a nicer demonstation than mine. Thank you.