UK 1998, Show that $hxyz$ is a perfect square. 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.  

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.
