In this thread:

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 $\displaystyle d|z$.