# 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

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 $g=1$.

• I assume you mean $h(y-x)$ is a square, as $h(x-y)$ is negative. Hint: Look at $x-y=\frac{xy}{z}$. – deinst Jul 14 '14 at 22:54
• @deinst thank you for pointing out. It is true indeed, I will rectify the question. – Swadhin Jul 14 '14 at 22:55
• Study properties even and odd, see my post... – johannesvalks Jul 15 '14 at 0:15
• Possible duplicate of Show that $hxyz$ is a perfect square. It has more simpler solution than this approach. – Rezwan Arefin Feb 4 '17 at 21:36

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.

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.

• I wonder what does $p|a$ mean, $p$ is a dividor of $a$? – johannesvalks Jul 15 '14 at 0:57
• @johannesvalks Yes, I am using that meaning for this notation. – user84413 Jul 15 '14 at 1:04
• @user84414, ok - I am not familiar with that notation, it is a cute notation!!! – johannesvalks Jul 15 '14 at 1:06
• @user84414: However, when you say "Let $p|b-a$ with $p$ prime", you need to show that such a prime exists. You assume it. I believe that such a prime does not exists, for $b-a=1$, it has no divisors... We have solutions like $$\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\\\frac{1}{3} - \frac{1}{4} = \frac{1}{12},$$ and in all cases $b-a=1$. Perhaps I am missing something... – johannesvalks Jul 15 '14 at 1:07
• @johannesvalks You have a good point, so I should modify my answer to reflect this. (Perhaps I should try to prove that $b-a=1$, which would be better.) – user84413 Jul 15 '14 at 1:11

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.