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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.

Martin Sleziak
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Tithus248
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4 Answers4

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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.

Ross Millikan
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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

Bill Dubuque
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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 $\gcd(a, b) \neq 1$

Martin Sleziak
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John Sig
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  • The conclusion in the line "now as..." is unjustified. – Bill Dubuque Oct 05 '19 at 02:15
  • See the note. $d^{2}$ and $nm$ are two numbers with gcd 1, that have a square product. – John Sig Oct 05 '19 at 12:47
  • But $\ \gcd(d^2,nm) = 1\ $ is not generally true. Also the proof of the Note is not correct. – Bill Dubuque Oct 05 '19 at 12:50
  • $gcd(d^{2}, nm)=1$ is based on the fact that $gcd(d, n)=1$ and $gcd(d, m)=1$.

    Also why is the proof not correct? It is a proof by contradiction.

     – John Sig Oct 05 '19 at 13:07
  • But that's not true, e.g. consider $, 8(18) = 12^2,$ where $,m,n,d = 4,9,2,$ and $,(m,d) = (4,2) = 2 \neq 1.\ $ As for the Note, the argument not derive the claimed result. – Bill Dubuque Oct 05 '19 at 13:11
  • But $gcd(m, d) = 2$ in this case – John Sig Oct 05 '19 at 13:16
  • Thus $,(d^2,nm)\neq 1$ so your claim is false. – Bill Dubuque Oct 05 '19 at 13:18
  • No if $(d, m) = 1$ and $(d, n) = 1$ then $(d^{2}, nm)=1$, that's my argument. – John Sig Oct 05 '19 at 13:20
  • But $,(d,m) = 2,,$ not $1\ \ $ – Bill Dubuque Oct 05 '19 at 13:22
  • That's your example not mine. That's why it is not a contradiction to my argument. Also the note holds because it indicates that the two $a, b$ have distinct square prime factors and as so they are square. A counterexample would be: Find me three numbers $(n, m, d)$ such that if $(d, m) = 1$ and $(d, n) = 1$, then $(d^{2}, nm) \neq 1$ – John Sig Oct 05 '19 at 13:24
  • But your argument fails on that example, which is a particular case of the OP's problem. There are infinitely many more counterexamples like that to your argument. I can't make any sense of your argument in the note. It seems you have problems understanding logical deduction. – Bill Dubuque Oct 05 '19 at 13:27
  • So the problem with my argument was that I assumed that (d, m) = 1, in the second line. Now I removed that part from the proof and I think it still works. Also please don't be insulting, – John Sig Oct 05 '19 at 13:41
  • There are still errors, e.g.. $,x_1 x_2/ d^2 ,$ is the square of a rational but your note only works for squares of integers (so it seems, it is not clear what is intended there). We can complete the proof in various ways, e.g. see my answers here. – Bill Dubuque Oct 05 '19 at 14:34
  • If $d^{2} \vdash x_{1}x_{2}$ then $\frac{x_{1}x_{2}}{d^{2}}$ is not rational. It is an integer and it should be square, as dividing a square by a square that divides it should give you again a square. Also this is not related with my note. My note has to do with $(n, m) = 1$. – John Sig Oct 05 '19 at 14:53
  • Again it is not clear what you intend. You have $,(x_1/d)(x_2/d) d^2 = c^2.\ $ How do you proceed from that? – Bill Dubuque Oct 05 '19 at 15:05
  • As for the Note, perhaps you intend an inductive proof e.g. as here, using FTA = unique prime factoization, but so little is said I can't be sure. – Bill Dubuque Oct 05 '19 at 15:30
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    Yes it has to do with FTA. Also can you have $d^{2} \vdash c^{2}$ and $d \not \vdash c$? If no then $\frac{c}{d}$ is an integer and $nm = (\frac{c}{d})^{2}$ is a square. My argument is really simple: prove that n*m is a square and then prove that n, m are both squares. – John Sig Oct 05 '19 at 15:48
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    Yes, $,d^2\mid c^2,\Rightarrow, d\mid c,,$ which I proved by the Rational Root Test in one of my answers (it can also be proved by Euclid's Lemma, unique factorizaton, etc). It is an important step where errors are often made, so it should be explicitly mentioned and cited / proved if it is used. With the changes you propose it seems that what you now intend is equivalent to what I wrote in the linked answer (which is different from what you first wrote). But there stil are gaps (and doubts) on what is written in the answer – Bill Dubuque Oct 05 '19 at 16:39
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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

Bill Dubuque
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