Does n² has all the factors n has? How can I be sure?

[MonkeyKid]

First of all, pardon me for my poor English, which is even worse for mathematics.

As an answer to my question, I think it does have all the factors n has. But I can't write mathematical proof of it, and so I can't make that statement. It's probably easy. I can write that in the opposite direction though:

let a, b, c, ..., z be the factors of N:

(a * b * c * ... * z) = N

then

(a * b * c * ... * z)² = N²
(pretty simple thinking)

but I can't write that in the other direction... I mean, beginning with N² and it's factors, conclude that it has at least all the factors N has.

 

[jbunniii]

Well, let's write $N = a_1 a_2 \ldots a_n$ and choose any factor, say $a_1$. Then $N^2 = (a_1 a_2 \ldots a_n)(a_1 a_2 \ldots a_n) = a_1*X$ where $X = (a_2 a_3 \ldots a_n)*N$. Therefore $a_1$ is a factor of $N^2$.

[edited] to use better notation

 

[MonkeyKid]

Well, let's write $N = a_1 a_2 \ldots a_n$ and choose any factor, say $a_1$. Then $N^2 = (a_1 a_2 \ldots a_n)(a_1 a_2 \ldots a_n) = a_1*X$ where $X = (a_2 a_3 \ldots a_n)*N$. Therefore $a_1$ is a factor of $N^2$.

[edited] to use better notation

Thank you, it took me a while to understand, because I forgot I was talking about prime factors. In fact I was, and the title should be different. My mistake. I'm glad you saw through it. That's a neat explanation by the way. I like math more and more with each passing day.