1. ## divisibility proof

http://www.mathhelpforum.com/math-he...an-triple.html

I was trying to prove that if d^2 divides z^2, then d divides z.

It seems like it should be an easy proof, but the proof I came up with is long and unclear (maybe even wrong). Can anyone come up with a better proof?

Here it is:
By definition, $\displaystyle d^2|z^2$ means there is a positive integer n such that $\displaystyle z^2=nd^2$. Suppose n is not the square of an integer. Let $\displaystyle n=ms^2$, where m>1 has no square divisors. Let p be a prime divisor of m. We have $\displaystyle z^2 = nd^2 = ms^2d^2$, and since the powers of p in $\displaystyle z^2$, $\displaystyle s^2$, and $\displaystyle d^2$ are even, the power of p in m must be even. But then m has a square factor. By contradiction, therefore, n must be the square of an integer. Let $\displaystyle n=a^2$. Then $\displaystyle z^2=a^2d^2$ and since z and d are positive and you can choose a to be positive, z=ad, which is the definition of $\displaystyle d|z$.

2. It'll be easier if you think in terms of prime factorizations, i.e. if

$\displaystyle d=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$ and $\displaystyle z=q_1^{r_1}q_2^{r_2}\cdots q_m^{r_m}$, with all p's and q's prime, and all s's, r's positive integers,

how would you express the relationships $\displaystyle d\mid z$ and $\displaystyle d^2\mid z^2$?

3. So:
$\displaystyle d|z$ means that if $\displaystyle p_i=q_j$ then $\displaystyle s_i\le{r_j}$
and:
$\displaystyle d^2|z^2$ means that if $\displaystyle p_i=q_j$ then $\displaystyle 2s_i\le{2r_j}$
which are obviously equivalent statements.

I was looking for something more like the proof of $\displaystyle a|b\text{ and }a|c \rightarrow a|(mb+nc)$ for any integers m,n. From the definition of divisibility, there are integers r,s such that b=ra and c=sa. Then mb+nc = (mr+ns)a and so a divides mb+nc.

Maybe what I'm looking for doesn't exist...