Let's show that \[a^2|z^2\Longrightarrow a|z\]

• Let \(d=\gcd(a,z)\).
• Then we can write \(z^2=a^2 \cdot k\) for some integer \(k\), and immediately write \[(z')^2 d^2=(a')^2 d^2 k\] for some integers \(z'\) and \(a'\), by definition of gcd. (That is, \(z=z'd\) and \(a=a'd\).)
• Cancelling the \(d^2\) (yes, we do assume this property of integers) yields \[(z')^2=(a')^2 k\; .\]
• Since \(gcd(a',z')=1\), we have \(a'x+z'y=1\) for some \(x,y\in\mathbb{Z}\); now we substitute for \(1\) in \(a'\cdot 1\cdot x\) (!) to get \[a'(a' x+z'y)x+z'y=1\]
• Now we have that \(a'^2 x+z'(a'xy+y)=1\), so that \(\gcd((a')^2,z')=1\) as well.
• But of course \(a'|(z')^2\). Clearly if a positive number is a divisor, but their greatest common divisor is 1, then that number is going to have to be 1 by definition of divisors. So \(a'=1\).
• (If \(a'\) was negative, the same argument for \(-a'\) shows \(-a'=1\), so really \(a'=\pm 1\).)
• Hence \(a=a'd=\pm d\), which is a divisor of \(z\), we have the desired result.

Let's show that if \(j^2=mn\) and \(\gcd(m,n)=1\), then \(m\) and \(n\) are also both perfect squares.

• First, we will need a general fact about gcds -- that \[\gcd(x,y)^2=\gcd(x^2,xy,y^2)\; .\] Can you prove this using the set of divisors definition of gcd, or using the definition \(\gcd(a,b,c)=\gcd(\gcd(a,b),c)\) and other things you already know?
• Then take \(m\). We know that \(1=\gcd(m,n)=\gcd(m,n,j)\).
• So \(m = m\gcd(m,n,j)=\gcd(m^2,mn,mj)=\gcd(m^2,j^2,mj)\).
• Now we use the fact, so that \(m=\gcd(m,j)^2\). That's a perfect square.
• The same argument with \(n\) and \(j\) yields \(n = \gcd(n,j)^2\).