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This problem appeared multiple times in MSE -

UK 1998, Show that $hxyz$ is a perfect square.
Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are perfect squares

They seem not to solve my problem. Also the posts are old enough. This problem appeared in UK MO 1998. I found this solution in the book "104 Number Theory Problems from USA Math Camp" written by Titu Andrescu - enter image description here
I don't understand the red boxed step. How we get
$c(b'-a') = a'b'g$ implies $g \mid c$?
why it cant occur that $g \mid (b'-a')$ and $a'b' \mid c$?
I don't understand from here till end.
Both of the linked question have answers with different approaches. But I want a solution which continues from here.

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It suffices to prove $\,abc\,$ and $\,b\!-\!a\,$ are squares, by $\ hxyz = h^4abc,\,$ $\, h(z\!-\!y) = h^2(b\!-\!a)$

$$\begin{align}{\rm multiplying}\ \ \ \dfrac{1}{c}\ &=\ \dfrac{1}a\ -\ \dfrac{1}b\ \ \ {\rm by}\ \ \ g := (a,b)\\[0.7em] {\rm yields}\ \ \ \dfrac{(a,b)}{c}\ &=\ \dfrac{1}{a'}\ -\ \dfrac{1}{b'}\ =\ \dfrac{b'-a'}{{a'b'}}\end{align}$$

Both fractions are reduced (i.e. in lowest terms) since $\,{((a,b),c)}=1=(b'\!-a',a'b'),\,$ therefore they have equal numerator and denominator (wlog $\ge 0),\,$ hence

$$b'\!-a' = (a,b)=g\ \Rightarrow\ b\!-\!a = g(b'\!-\!a') = g^2$$

$$ a'b' = c\ \Rightarrow\ abc = g^2 a'b'c = g^2 c^2 = (gc)^2$$

Remark $\ $ Notice how much clearer the argument becomes when presented in fractional form. See this post on unique fractionization for more on the uniqueness of reduced fractions.

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  • $\begingroup$ Great! I never seen this kind of solutions for this type of problems. Too simple ! :) $\endgroup$ – Rezwan Arefin Feb 5 '17 at 7:09
  • $\begingroup$ But didn't get how $\frac{b'-a'}{a'b'}$ implies $c= a'b'$? Explain a little bit more please $\endgroup$ – Rezwan Arefin Feb 5 '17 at 7:11
  • $\begingroup$ @Rezwan I added the full argument, and clarified the use of uniqueness of reduced fractions. $\endgroup$ – Bill Dubuque Feb 5 '17 at 15:38
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Good job asking a specific question—it makes it very easy to provide useful comments! And indeed, this step of the proof is incorrect: when $a=6$, $b=15$, and $c=10$, we have $\frac16-\frac1{15} = \frac1{10}$ and $g=\gcd(a,b)=3$ does not divide $c$.

(By the way, in your comments, you say "why can't it occur that $g\mid (b'-a')$...?" I want to remind you that $g\mid c(b'-a')$ does not necessarily mean that $g\mid c$ or $g\mid(b'-a')$; it could be that $g$ is the product of some divisor of $c$ times some divisor of $b'-a'$.)

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  • $\begingroup$ But the books is quite well known :| "104 Number Theory Problems from USA Math Camp" Written by Titu Anrescu Trainer of USA MO Team:| $\endgroup$ – Rezwan Arefin Feb 4 '17 at 22:12
  • $\begingroup$ Math truth > well known :) $\endgroup$ – Greg Martin Feb 4 '17 at 22:13
  • $\begingroup$ wait .. I think the statement is right ... $a,b,c$ are not any arbitrary integers. It must be true that $\frac{1}{x} - \frac{1}{y} = \frac{1}{z}$ and $x = ad, y = bd, z = cd$ with $d = gcd(x,y,z)$. $\endgroup$ – Rezwan Arefin Feb 4 '17 at 22:23
  • $\begingroup$ But if x = 6, y = 15 and z = 10 then d=gcd(x,y,z) = 1 so $x = a =6$ and $y = b = 15$ and $z = c = 10$. And the observation is valid. $\endgroup$ – fleablood Feb 4 '17 at 22:34
  • $\begingroup$ @GregMartin Now how to continue from here ? $\endgroup$ – Rezwan Arefin Feb 4 '17 at 22:35
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Actually I get the exact opposite as the text does.

As $\gcd(a,b,c) = 1$ and $\gcd(a,b) = g$ we know $\gcd(g,c) = 1$

So we have $c(b'-a') = a'b'g$ so $g|(b'-a')$ Furthermore we know $\gcd(a'-b',a') = \gcd(a'-b',b') = 1$ so $a'b'|c$.

So we have $\frac {c}{a'b'}(b'- a') = g$. But $\gcd(g,c) =1$ so $\frac{c}{a'b'} = 1$ so $g = b'-a'$ (but that's not relevant) and $c = a'b'$ (and that is relevant).

So $hxyz = h^4abc = h^4(ga')(gb')(a'b') = (h^2ga'b')^2$.

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If I were to do it entirely on my own:

Let $h = \gcd(x,y,z); x = hx'; y= hy'; z = hz'$

$hxyz = h^4(x'y'z)$ and $h(y-x) = h^2(y'-x')$ so $hxyz$ is a perfect square if and only if $x'y'z$ is and $h(y-x)$ is a perfect square if and only if $y'-x'$ is. So wolog we may assume $h= 1$.

$\frac 1z = \frac 1x - \frac 1y$

$\frac {xy}z = y -x$. Presumable $y \ne x$ if is did we'd have $z = 0$ and $hxyz = 0^2$ and $h(y-x) = 0$ is trivial. So presume $y \ne x$

$z = \frac {xy}{y-x}$

So $y-x|xy$.

Let $g=\gcd(x,y)$ and $x=gx';y=gy'$ and $\gcd(x',y') =1$.

$y'-x'|gx'y'$. If $p$ prime is a factor of $x'$ then $p\not \mid y'$ and $p|y'-x'$. Likewise for any prime factor of $y'$. So $\gcd(y' -x',x'y') = 1$ and $y'-x'|g$. Let $y'-x' = kg$. And as $\gcd(y'-x', x'y') =1$ we know $\gcd(k, x'y') = 1$.

So $z = \frac {xy}{y-x} = \frac{gx'y'}{y'-x'} = \frac{x'y'}k$. But $\gcd(x'y',k) = 1$ so $k = 1$ and $y' - x' = g$

So that proves $y-x = g(y'-x') = g^2$ is a perfect square.

So $xyz = xy\frac{xy}{y-x} = xy\frac{xy}{g^2}=(\frac{xy}{g})^2 = (x'y)^2 = (xy')^2 = (gx'y')^2$ is a perfect square.

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  • $\begingroup$ Why $g\mid (b'-a')$? $\endgroup$ – Rezwan Arefin Feb 4 '17 at 23:02
  • $\begingroup$ Because $g$ is relatively prime to $c$. $\endgroup$ – fleablood Feb 4 '17 at 23:03
  • $\begingroup$ If $\gcd(m,n) = 1$ and $n|mx$ then $n|x$ because $m$ and $n$ have no factors in common. $\endgroup$ – fleablood Feb 4 '17 at 23:05
  • $\begingroup$ I feel like I might be wrong though. In the a=6, b=15, c = 10 we have $a'=2;b'=5$ so $c' =a'b'=10$. Can anyone think of a case where $c \ne a'b'$? I don't see my error though. And we do know that $\gcd(a,b,c) = \gcd(\gcd(a,b),c)$ correct? $\endgroup$ – fleablood Feb 4 '17 at 23:09
  • $\begingroup$ And I do find it odd that a well reputed book would be just plain wrong. I get that $g \not \mid c$ unless $g = 1$. That's a huge error. I'm not ... comfortable with saying of course I'm right and the book is wrong but... I don't see that I am. And Greg Martin's counter example... counters. $\endgroup$ – fleablood Feb 4 '17 at 23:12

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