Number Theory: In Context and Interactive

First, let \(d=\gcd(a,z)\text{.}\) Then we can write \(z^2=a^2 \cdot k\) for some integer \(k\text{,}\) and immediately write

for some integers \(z'\) and \(a'\text{,}\) by definition of gcd. (That is, \(z=z'd\) and \(a=a'd\text{.}\) Also note that \(z',a'\) are now relatively prime; it is not hard to prove using the techniques of the previous chapter, or see Exercise 6.6.7.)

Cancelling the \(d^2\) (yes, we do assume this property of integers) yields

Since \(\gcd(a',z')=1\text{,}\) we have \(a'x+z'y=1\) for some \(x,y\in\mathbb{Z}\text{;}\) now we substitute for \(1\) in \(a'\cdot 1\cdot x\) (!) to get

Now we have that \(a'^2 x^2+z'(a'xy+y)=1\text{,}\) so that \(\gcd((a')^2,z')=1\) as well. But of course \(a'\mid (z')^2\text{.}\) 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 \(a'=1\text{.}\) (If \(a'\) was negative, the same argument for \(-a'\) shows \(-a'=1\text{,}\) so really \(a'=\pm 1\text{.}\))

Hence \(a=a'd=\pm d\text{,}\) which is a divisor of \(z\text{,}\) we have the desired result.

First, we will need a general fact about gcds:

See Exercise 3.6.22.

We know that \(1=\gcd(m,n)=\gcd(m,n,j)\text{,}\) so

Now we use the fact, so that

That's a perfect square.

The same argument with \(n\) and \(j\) yields \(n = \gcd(n,j)^2\text{.}\)