Hey, We can trivially find $z=0$ when $x<0$ and $y>0$. So without loss of generality (noting that if $p \in M$ then $-p \in M$) we can work with $y>x>0$ Consider the set, $N$, of squares of elements of $M$. $N$ contains non-negative rational numbers of the form $1 + \frac{2mn}{m^2+n^2}$. If we can find any $q \in N$ satisfying $x^2<q<y^2$ then the positive square root of $q$ will give an appropriate value of $z$. But $1 + \frac{2mn}{m^2+n^2} = 1 + \frac{2}{\frac{m}{n}+\frac{n}{m}} = 1 + \frac{2}{r+r^{-1}}$ for any rational $r$. Consider $f(r) = 1 + \frac{2}{r+r^{-1}}$ as a real function. $r + r^{-1} = 0$ has no real solutions. Only discontinuity value of the function is at $z = 0$, but on both sides of $r=0$ the function tends to $1$ continuously, so we could define $f(0)=1$ to produce an entirely continuous function. Suppose $f(r_1) = x^2, f(r_2) = y^2$. then by continuity, we can always find a sub-interval $[a, b]$ of $[r_1,r_2]$ (the square bracket denotes the closed set, we don't require $r_1 < r_2$ in this argument) such that whenever $c \in [a,b]$ then $f(c) \in [f(z_1),f(z_2)]$. Intuitively, we can always find a monatonic interval with the given bounds. But since the rationals are dense in the reals, we can certainly choose $c$ to be a rational $\in [a,b]$. Take $z =\sqrt{f(c)}$. Woop.