# Math contest proof problem fractions

Could someone help me with this?

Let $x, y, z$ be positive integers with greatest common divisor $1$. If $\frac 1 x +\frac 1 y=\frac 1 z$, then show that $\sqrt{x + y}$ is an integer.

$x+y = \frac{xy}{z}$ is an integer, so $z \mid xy$.
Let $z = z_x \times z_y$ where $z_x \mid z$ and $z_y \mid y$.
Let $x = z_x \times x_x$ and $y = z_y \times y_y$, where every variable is an integer.

Then we have that $z_xx_x + z_y y_y = x_xy_y$

Apply the condition that $\gcd(x,y,z)=1$ to conclude that $x_x | y_y$ and $y_y | x_x$. (Fill in this argument. If you want a further hint ...

$\gcd(x,y,z) = 1 \Rightarrow \gcd(x_x, z_y ) = 1$. Since $x_x(z_x-x_y) = -z_y y_y$ hence $x_x \mid y_y$

Hence $\sqrt{x+y} = x_x$ is an integer.

• Thank you! Okay, I'll do my best to fill in the rest, but give me a hint as to how the relative primality plays a role. – Yadnarav3 Aug 31 '13 at 2:04
• @user87611 If you scroll over they grey box, you will see the step. – Calvin Lin Aug 31 '13 at 4:04

Note: we can replace the condition $$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$ with the following weaker one:

For every prime $p$, $$\nu_p(x+y)=\nu_p(x)+\nu_p(y)-\nu_p(z)=S_p\geq 0$$

First, suppose that $\nu_p(z)=0$. Then $\nu_p(x+y)=\nu_p(x)+\nu_p(y)$, so $\nu_p(x)=\nu_p(y)$ and $S_p$ is even.

Otherwise, suppose wlog that $\nu_p(x)=0$. If $\nu_p(y)=0$, then $S_p=0$, so suppose $\nu_p(y)\geq 1$. Then $\nu_p(x+y)=0\implies S_p=0$ again.

Thus in all cases, $S_p$ is even. It follows that $x+y$ is a perfect square.

Write $y=n-x$, so that what we want to show is that $n$ is a square. Note that gcd-wise, $(x,n,z)=(x,y,z)=1$. Now let $d=(x,n)$ and write $n=dm$ and $x=du$, with $(m,u)=1$. Note that $(d,z)=1$ as well. The equation ${1\over x}+{1\over n-x}={1\over z}$ rewrites as $nz=x(n-x)$, which becomes

$$dmz=du(dm-du)$$

Cancelling a $d$ leaves $mz=du(m-u)$, and this implies $d|m$, since $(d,z)=1$, so writing $m=d\ell$, we get (after cancelling the $d$)

$$\ell z= u(d\ell-u)$$

This implies $\ell|u^2$. But $\ell$ and $u$ are relatively prime, hence $\ell=1$, from which we conclude that $n=d^2$.