Let p be the smallest prime dividing a positive integer n with n > p. Prove n is the product of two primes if p^3 > n.

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- Nov 11th 2009, 11:14 AMAwesomeDesiKidProoving Primes
Let p be the smallest prime dividing a positive integer n with n > p. Prove n is the product of two primes if p^3 > n.

- Nov 11th 2009, 02:30 PMtonio

If you meant "the product of two DIFFERENT primes", and I can't see what else could you mean, then this is false: $\displaystyle p=2$ is the smallest prime dividing $\displaystyle n=4\,\,and\,\,p^3=2^3=8>4=2\,,\,\,yet\,\,n$ is not the product of two (different) primes.

Tonio - Nov 11th 2009, 03:02 PMAwesomeDesiKid
well, thats what i was confused about...i wanted to know if there could be any other meaning for it

- Nov 11th 2009, 03:09 PMAwesomeDesiKid
no, the two primes can be same...still confused....(Headbang)

- Nov 11th 2009, 03:24 PMtonio

That's what I was afraid: an ill-posed, or trivially-posed, problem: as p < n AND p is one of the primes (the lesser one, btw) that divides n, then it is CLEAR that it must be more than merely p that divides n...! So if the question's going to be meaningful, it MUST be that it meant to ask that the primes are different...but then, as seen, the claim is false. (Happy)

I think the question could be asked as follows: let n be a natural number which is not the power of a prime number and such that p is the minimal prime dividing it. Then, if p^3 > n then p is the product of two different primes.

But then the claimis very easy: as n is not the power of a prime it is divided by at least two primes. If n were divided by three primes p < q < r, then p^3 < p*q*r <= n, so that p^3 > n isn't true, contradiction ==> n is divided by two different primes and we're done.

Tonio