Two integers whose product is square

Question

I am trying to prove the following statement.

Let $d = \gcd(x_1,x_2)$. If $x_1 x_2$ is a square, then ($x_1 = d M^{2}$ and $x_2 = dN^2$) or ($x_1 = -d M^{2}$ and $x_2 = -dN^2$), where $\gcd(N,M) =1$.

This is true if $d=1$: If a and b are relatively prime and ab is a square, then a and b are squares.

Thank you very much.

Answer 1

Note $\,x_1 x_2 = c^2\,\Rightarrow\, x_1/x_2 = (c/x_2)^2 = (a/b)^2\,$ for $\,a/b\,$ in least terms, i.e. $(a,b) = 1$.

By Euclid $\begin{align} (a,\ b)\ &=1\\ \iff\! (a^2,b^2)&=1\end{align}\,\ $ so $\,\ \dfrac{x_1}{x_2} = \dfrac{a^2}{b^2}\,\Rightarrow\begin{align}x_1 = d a^2\\ x_2 = d b^2\end{align},\ d\in\Bbb Z\ $ by Unique Fractionization

Answer 2

Here is an informal sketch: Consider a prime $p$ which divides $x_1x_2$. Because $x_1x_2$ is a square, the maximum power of $p$ dividing $x_1x_2$ is even, call it $2k$. The maximum power of $p$ dividing $d$ is some number $m$, which is the maximum power dividing one of $x_1,x_2$. Say it is $x_1$. Then the maximum power of $p$ divding $x_2$ is $2k-m$, $p$ does not divide $M$ and the power of $p$ dividing $\frac {x_2}d$ is $2k-2m$, which is even. This justifies that $\frac {x_2}{d}$ is a square.

Answer 3

As $\gcd(x_{1}, x_{2}) = d$, there exists an $n, m$, such that:

$x_{1} = \pm \;d\cdot m$ with $m \geq 1$

and

$x_{2} = \pm \;d\cdot n$, with $n\geq 1$.

Their sign should be the same as a square is positive.

Now as $d^{2} \vdash x_{1}x_{2}$, $nm = \frac{x_{1}x_{2}}{d^{2}}$ should be also square.

Also as $\gcd(x_{1}, x_{2}) = d$, $\gcd(\frac{x_{1}}{d}, \frac{x_{2}}{d})=1$, that is $\gcd(n, m) = 1$, $n$ and $m$ should be also square.

This means there exists an $N\geq 1$ and an $M\geq 1$, such that: $n = N^{2}$ and $m = M^{2}$.

Note

To prove that:

if two numbers $a, b$, with $\gcd(a, b) = 1$ have a square product each of them should be a square

you can say that:

if there exists a prime number such that $p^{2k} \vdash ab$, then either $p^{2k} \vdash a $ or $p^{2k} \vdash b$, because otherwise $\gcd(a, b) \neq 1$

Answer 4

Define $y_i = x_i /d.\,$ Then $\,x_2x_2 = k^2\,\Rightarrow\,y_1 y_2 = (k/d)^2\,$ and $\,k/d\in\Bbb Z\,$ by the Rational Test. Since $y_1,y_2$ are coprime factors of a square it follows that both are squares (up to sign). $ $ QED