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Problem: [UK 1998] Let $x,y$ and $z$ be positive integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{1}{z}.$$ Let $h=(x,y,z).$ Prove that $hxyz$ is a perfect square.

My Attempt: Of course $y>x$ and therefore let $y=x+a$ for some $a>0.$ Then $$z=\frac{x(x+a)}{a}.$$ Also $$h=(x,x+a,z)=(x,a,z).$$ Now $a|x(x+a)$ since $z$ is a positive integer. Let $p$ be a prime divisor of $a$ then $p|x$ or $p|x+a$. But if $p|x\Rightarrow p|x+a$ and similarily if $p|x+a\Rightarrow p|x.$ In any case we deduce that $a|x$ and $a|x+a$ also. Thus $$z=\frac{x(x+a)}{a}=\frac{a^2k_1}{a}=ak_1$$ and $x=ak_2.$ Thus $h=a.$ The expression $hxyz=a(az)(z)=(az)^2$ which is indeed a whole square.

I would like to know whether this proof is correct or not since the argument provided in the solution is quite different from the one I've written in this question.

Shrey Aryan
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    Does $(x,y,z)$ mean their highest common factor (greatest common divisor)? – Henry Nov 22 '16 at 08:37
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    "In any case we deduce that a|x " Actually you can't because you only considered prime factors and not powers of prime factors. Say a = 25=5^2 and x= 15. then 25|15(15 + 25) but 25 doesn't divide 15. – fleablood Nov 22 '16 at 08:37
  • @fleablood: very neat: so you have $\dfrac{1}{15}-\dfrac{1}{40}=\dfrac{1}{24}$ – Henry Nov 22 '16 at 08:47
  • I second @fleablood, you appear to prove only that all prime divisors of $a$ divide $x$. – Andreas Caranti Nov 22 '16 at 09:15
  • In fact is a duplicate: http://math.stackexchange.com/questions/867392/let-x-y-z-be-positive-integers-such-that-frac1x-frac1y-frac1z – iam_agf Nov 22 '16 at 09:40
  • @MonsieurGalois, the question has appeared several times on MSE. But here we are discussing a possibly novel proof. – Andreas Caranti Nov 22 '16 at 09:59
  • @Henry that was me being lucky. I chose 15 and 25 arbitrarily I could have chosen 75 and 35 and 1/35 - 1/110 = (22 - 7)/110 = 15/110 = 3/22 != 1/n. I just wanted to show the fallacy in the argument. – fleablood Nov 22 '16 at 17:01

2 Answers2

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Let $z=hc, x=ha,y=hb$ so $(a,b,c)=1$. Notice that:

\begin{align*} hxyz=h^4abc \end{align*}

We need to see that $abc$ is an square. Notice that from the equation, \begin{align*} z=\frac{xy}{y-x}&=\frac{h^2ab}{h(b-a)}=\frac{hab}{b-a}\\ c&=\frac{ab}{b-a}\\ \Rightarrow ab=c(b-a) \end{align*}

So from the first equation: $$hxyz= (h^4c)(c(b-a))=h^2c^2(b-a)$$

So we need to check that $b-a$ is an square. From the last equation, $c(b-a)=ab$, if $d\mid b-a$ is a prime, then $d\mid ab$, so $d\mid a$ or $b$, but any of them lets that $d$ divides the other, so $d^2\mid ab$. Since $(a,b,c)=1$, as we have $c(b-a)=ab$, $d^2\mid c(b-a)$, but $(d,c)=1$, so $d^2\mid (b-a)$, implying that for each prime divisor of $b-a$, there are two. If there is no $d$ prime such that $d\mid b-a$, then $b-a=1$, that is a square too. Then you have that

$$hxyz=h^4c^2(b-a)$$

Is really an square.

iam_agf
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    Your last paragraph shows that the prime factorization of $b-a$ entails powers that are greater than or equal to $2$. It doesn't show that all powers are even. A fix only requires a slight rewriting of your argument. – yurnero Nov 22 '16 at 17:47
  • Thanks, let me correct it. – iam_agf Nov 23 '16 at 00:26
  • Ok, I don't know how to avoid that problem. If you know it, please explain it to me. – iam_agf Nov 23 '16 at 01:17
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    Suppose that $p^{2k+1}|b-a$ and $p^{2k+2}\not|b-a$. Then, because $p^{2k+1}|ab$, either $p^{k+1}|a$ or $p^{k+1}|b$. But if $p^{k+1}|a$ then of course $p^{k+1}|b$ and similarly if $p^{k+1}|b$ then $p^{k+1}|a$. Thus, $p^{k+1}$ divides both $a$ and $b$. And so $p^{2k+2}$ divides $ab$. But since $p^{2k+2}\not|b-a$ this means $p$ is a common factor among $a,b,c$. Contradiction. – yurnero Nov 23 '16 at 01:52
  • How we know that $(d,c)=1$ ??? sorry but i did not understand this step, we just assume that there is a prime such that it divides $b-a$ but does not divide c , but in this case how we know that such a prime exist ??? or is there something else in this that i am missing ...??? – Ishan May 18 '20 at 12:13
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Let's try your argument with $x = 15; y = 40; z = 24$ where $\frac 1{15} - \frac 1{40}= \frac{8}{8*15} - \frac{3}{3*40} = \frac 5{5*24} = \frac 1 {24}$ and $h = (15,40,24) = 1$ and $hxyz = 1*3*5*8*5 *2*8 = (3*5*8)^2$.

You argue $40 > 15$ and $a = 40 -15 = 25$.

So $z = 24 = \frac {15*(15 + 25)}{25}$ which is indeed true.

You argue that $h = (15,40,24) = (15,25,24)$ which is true.

You argue that if $p = 5$ a prime factor of $a = 25$ than $5|15*(15+ 25)$ and that as $5|25$ then $5|x(x + 25) \iff 5|x = 15$ which is true.

Then you conclude that since $5|15$ then $25|15$ which is not true. We can only conclude the single power products of the prime factors divide $x$; not any powers of any prime.

You argue $z = \frac{x(x+a)}{a} = \frac {a^2k_1}{a} = ak^2$ or $24 = \frac{15(15 + 25)}{25} = \frac {25^2*\frac 35(\frac 35 +1)}{25} = 25*\frac 35\frac 85 = 25*\frac {24}{25}$. In which case $k_1=\frac {24}{25}$ is not an integer.

Likewise $15 = 25*\frac 35$ but $\frac 35$ is not an integer. $25$ is not a common divisor to $15, 24, $ and $h \ne 25$.

Actually $h = a > 1$ should have set off huge red flags. $h = (x,y,z)$ so this means $x,y,z$ are never relatively prime? Well, what happens if we factor $h$ out of them all? We have $\frac 1{x'} + \frac 1{y'} = \frac 1{z'}$ where $y' = x + 1$ and $z' = x'(x'+1)$? Is that really the only possibility?

I think it can be fixed but it doesn't work as is.

fleablood
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