Proposition 3.7.1. When perfect squares divide each other.
For integers \(a,z\) it is true that
\begin{equation*}
a^2\mid z^2\Longrightarrow a\mid z
\end{equation*}
Section 3.7 Two facts from the gcd
Here are two facts that seem really obvious but do need proofs. All can be done just with the gcd, using no facts about primes from Chapter 6 as would typically be done. Kudos go to users Math Gems and coffeemath at math.stackexchange.com for most of these clever arguments. See this question for Proposition 3.7.1 and this question for Proposition 3.7.2.
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math.stackexchange.com/users/23500/math-gems16
math.stackexchange.com/users/30316/coffeemath17
math.stackexchange.com/users/18
math.stackexchange.com/questions/286099/19
math.stackexchange.com/questions/286101/Proof.
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
\begin{equation*}
(z')^2 d^2=(a')^2 d^2 k
\end{equation*}
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
\begin{equation*}
(z')^2=(a')^2 k\text{.}
\end{equation*}
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
\begin{equation*}
a'(a' x+z'y)x+z'y=1
\end{equation*}
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.
Proposition 3.7.2. When the product of coprime numbers is a square.
If we have integers \(m,n,j\) such that \(mn=j^2\) and \(\gcd(m,n)=1\text{,}\) then \(m\) and \(n\) are also both perfect squares.
Proof.
First, we will need a general fact about gcds:
\begin{equation*}
\gcd(x,y)^2=\gcd(x^2,xy,y^2)
\end{equation*}
See Exercise 3.6.22.
We know that \(1=\gcd(m,n)=\gcd(m,n,j)\text{,}\) so
\begin{equation*}
m = m\cdot \gcd(m,n,j)=\gcd(m^2,mn,mj)=\gcd(m^2,j^2,mj)
\end{equation*}
Now we use the fact, so that
\begin{equation*}
m=\gcd(m,j)^2\text{.}
\end{equation*}
That’s a perfect square.
The same argument with \(n\) and \(j\) yields \(n = \gcd(n,j)^2\text{.}\)
(For more ‘traditional’ proofs, see Section 6.4.)
