prove if n is square-free and n|m^2, then n divides m

[tenshineko]

Prove that if n is square-free and n divides m[sup:1aapn8cb]2[/sup:1aapn8cb], then n must divide m.

i'm not sure how to go about even starting this.
any help would be appreciated.
thanks

 

[daon]

Re: prove n divides m

What are you able to use? My suggestion: Try writing n, m as a product of primes grouped into powers (n will be a product of unique primes of power 1...why?). m^2 must contain them all in its factor tree, now what about m?

 

[tenshineko]

Re: prove n divides m

n will be a product of unique primes of power 1...why?

that's the square-free part...
um..

if i say n = p[sub:1xqi3i5o]1[/sub:1xqi3i5o] x p[sub:1xqi3i5o]2[/sub:1xqi3i5o] x p[sub:1xqi3i5o]3[/sub:1xqi3i5o]

and n | m[sup:1xqi3i5o]2[/sup:1xqi3i5o], so m[sup:1xqi3i5o]2[/sup:1xqi3i5o] = p[sub:1xqi3i5o]1[/sub:1xqi3i5o][sup:1xqi3i5o]x[/sup:1xqi3i5o] x p[sub:1xqi3i5o]2[/sub:1xqi3i5o][sup:1xqi3i5o]y[/sup:1xqi3i5o] x p[sub:1xqi3i5o]3[/sub:1xqi3i5o][sup:1xqi3i5o]z[/sup:1xqi3i5o]

is that the way i should be thinking about it?

 

[daon]

Re: prove n divides m

Kind of, but to make things more clear:

\(\displaystyle n=p_1p_2 \cdots p_k\)

\(\displaystyle m=q_1^{a_1}q_2^{a_2} \cdots q_t^{a_t}\)

\(\displaystyle m^2=q_1^{2a_1}q_2^{2a_2} \cdots q_t^{2a_t}\)

What you wrote is not really correct because in addition to the primes in n, m^2 can have more, different ones.

 

[PAULK]

Prove that if n is square-free and n divides m[sup:9bfoavlr]2[/sup:9bfoavlr], then n must divide m.

i'm not sure how to go about even starting this.
any help would be appreciated.
thanks

Perhaps you start by noting that:

1. n is a product of primes, none of which appears as a factor more than once.
2. If p1 is one of them, then if n | m^2, then p1 | m^2, from which you conclude p1 | m.
3. If p1 and p2 each | m, then p1 p2 | m.