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

Question

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 both perfect squares.

My attempt: Let $x=ha,y=xb,z=xc$, then $a,b,c$ are positive integers such that $\gcd(a,b,c)=1$.
Now, suppose $\gcd(a,b)=g$, so, $a=ga',b=gb'$, where $\gcd(a',b')=\gcd(a',a'-b')=\gcd(b',a'-b')=1$. So, we have $c(b'-a')=ga'b'$, so $g\mid c$, and $g=1$.

Now, I do not know what to do. Please help. Thank you.

Answer 1

Let us consider

$$ \frac{1}{x} - \frac{1}{y} = \frac{1}{z}, $$

and

$$ h = \textrm{gcd}(x,y,z). $$

Say that

$$ \begin{eqnarray} x &=& h x',\\ y &=& h y',\\ z &=& h z',\\ \end{eqnarray} $$

then we obtain

$$ \frac{1}{x'} - \frac{1}{y'} = \frac{1}{z'}, $$

where

$$ \textrm{gcd}(x',y',z') = 1. $$

---

We can write

$$ \Big(y'-x'\Big)z' = x'y'. $$

Let us look to odd and even:

$$ \begin{array}{ccccl} x' & y' & z'\\ \textrm{even} & \textrm{even} & \textrm{even} & \Rightarrow & \textrm{gcd}(x',y',z') > 1.\\ \textrm{odd} & \textrm{odd} & \textrm{odd} & \Rightarrow& \textrm{contradiction as $xy$ is odd but $(x-y)z$ is even}.\\ \textrm{odd} & \textrm{even} & \textrm{even}\\ \textrm{even} & \textrm{odd} & \textrm{even} \end{array} $$

So we write

$$ \begin{eqnarray} y' &=& x' + 2p + 1,\\ z' &=& 2q, \end{eqnarray} $$

whene

$$ x'^2 + \Big(2p+1\Big)x' = \Big(2p+1\Big)2q, $$

so

$$ x' = - \frac{2p+1}{2} + \frac{2p+1}{2} \sqrt{1 + \frac{8q}{2p+1}}, $$

meaning that

$$ q = \frac{1}{2} r \Big( r + 1 \Big) \Big( 2p + 1 \Big), $$

so that

$$ \begin{eqnarray} x' &=& r \Big( 2p + 1 \Big),\\ y' &=& \Big( r + 1 \Big) \Big( 2p + 1 \Big),\\ z' &=& r \Big( r + 1 \Big) \Big( 2p + 1 \Big).\\ \end{eqnarray} $$

However $\textrm{gcd}(x',y',z') = 1$, thus $2p+1=1$, whence

$$ \begin{eqnarray} x &=& h r,\\ y &=& h \Big( r + 1),\\ z &=& h r \Big( r + 1 \Big).\\ \end{eqnarray} $$

---

It is clear that

$$ \begin{eqnarray} hxyz &=& \left[ h^2 r \Big( r + 1 \Big) \right]^2,\\ h(y-x) &=& \Big[ h r \Big]^2,\\ \end{eqnarray} $$

thus both $hxyz$ and $h(y-x)$ are perfect squares.

Answer 2

From @OP, we have $c(b-a) = ab$.

Suppose $gcd(a,b) = g$ and we write $a = ga'$ and $b = gb'$ and $g \ge h$

We have $c(b'-a') = ga'b'$. We have $gcd(c, a'b') = 1$.

So $c$ has to divide $g$. If $g > c$, then $gcd(c,g) \gt h$. Hence $c = g = h$

$\implies b' - a' = a'b' \implies a' \mid b'$, a contradiction since $gcd(a',b') = 1 \implies g = gcd(a,b) = 1$.

We now have $c(b-a) = ab$ where $gcd(a,b) = gcd(ab, b-a) = 1$

So $c \mid ab$. If $ab = kc$ then this contradicts $gcd(ab, b-a) = 1$ for $k \gt 1 \implies k = 1$

So $c = ab$. Hence $abc$ is a square. Also $b-a = 1$ which is a square.

Answer 3

As above, let $x=ha$, $y=hb$, and $z=hc$ where $h=gcd(x,y,z)$

Then $(y-x)z=xy\implies h^2(b-a)c=h^2(ab)\implies(b-a)c=ab$ with $gcd(a,b,c)=1$.

Let $p\vert b-a$ with $p$ prime. Then $p\vert ab\implies p\vert a$ or $p\vert b$; and

$\textbf{1)}$ if $p\vert a$, then $p\vert a$ and $p\vert b-a\implies p\vert b\implies p^2\vert ab\implies p^2\vert (b-a)c$ with $(p^2, c)=1\implies p^2\vert b-a$.

$\textbf{2)}$ Similarly, if $p\vert b$, then $p^2\vert b-a$.

[Notice that if $b-a$ has no prime divisor, then $b-a=1$.]

Thus $b-a$ is a perfect square, so it follows that

$hxyz=h^4abc=h^4(b-a)c^2$ and $h(y-x)=h^2(b-a)$ are perfect squares.