$\frac{1}{x}-\frac{1}{y} = \frac{1}{z}, h = gcd(x,y,z)$. Prove that $hxyz$ and $h(y-x)$ both are perfect squares.

Question

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 -
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.

score 1 · Answer 1 · answered Feb 04 '17 at 22:08

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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 $b'-a'$.)

answered Feb 04 '17 at 22:08

Greg Martin

78,820

But the books is quite well known :| "104 Number Theory Problems from USA Math Camp" Written by Titu Anrescu Trainer of USA MO Team:| – Rezwan Arefin Feb 04 '17 at 22:12
Math truth > well known :) – Greg Martin Feb 04 '17 at 22:13
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 $d = gcd(x,y,z)$. – Rezwan Arefin Feb 04 '17 at 22:23
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. – fleablood Feb 04 '17 at 22:34
@GregMartin Now how to continue from here ? – Rezwan Arefin Feb 04 '17 at 22:35

fleablood · Answer 2 · 2017-02-05T00:20:24.670

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

=====

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.

edited Feb 05 '17 at 00:20

answered Feb 04 '17 at 23:01

fleablood

124,253

Why $g\mid (b'-a')$? – Rezwan Arefin Feb 04 '17 at 23:02
Because $g$ is relatively prime to $c$. – fleablood Feb 04 '17 at 23:03
If $\gcd(m,n) = 1$ and $n|mx$ then $n|x$ because $m$ and $n$ have no factors in common. – fleablood Feb 04 '17 at 23:05
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? – fleablood Feb 04 '17 at 23:09
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. – fleablood Feb 04 '17 at 23:12
Then how to prove this? – Rezwan Arefin Feb 04 '17 at 23:13
I think I have proven it. I'm just ... uncomfortable. – fleablood Feb 04 '17 at 23:14
First linked question has an unaccepted answer... That's first part is really easy. but the second proof is missing something – Rezwan Arefin Feb 04 '17 at 23:16
@Rezwan The argument is much clearer when presented in fractional form - see my answer. – Bill Dubuque Feb 04 '17 at 23:36

score 1 · Accepted Answer · edited Apr 13 '17 at 12:19

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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.

edited Apr 13 '17 at 12:19

Community

1

answered Feb 04 '17 at 23:29

Bill Dubuque

272,048

Great! I never seen this kind of solutions for this type of problems. Too simple ! :) – Rezwan Arefin Feb 05 '17 at 07:09
But didn't get how $\frac{b'-a'}{a'b'}$ implies $c= a'b'$? Explain a little bit more please – Rezwan Arefin Feb 05 '17 at 07:11
@Rezwan I added the full argument, and clarified the use of uniqueness of reduced fractions. – Bill Dubuque Feb 05 '17 at 15:38

$\frac{1}{x}-\frac{1}{y} = \frac{1}{z}, h = gcd(x,y,z)$. Prove that $hxyz$ and $h(y-x)$ both are perfect squares.

3 Answers3