It should be centered at the origin and have the following vertices:\[[a,b,a^2-b^2], [-a,-b,a^2-b^2], [-b, a, -a^2+b^2], [b, -a, -a^2+b^2].\]We want all quadrances to be the same. By symmetry, only one equation is needed:\[4a^2+4b^2=(a-b)^2+(a+b)^2+4(a^2-b^2)^2\]or equivalently\[a^2+b^2=2(a^2-b^2)^2.\]
Our approach is to try to solve for one of the variables, say \(a\) in terms of \(b\). We will plug in different rational \(b\) and filter out those which give rational \(a\). A major insight is that filtering can be done by parametrization, or a change of variables if you wish. Let's start with completing the square:\[\left(a^2-\frac{4b^2+1}4\right)^2=b^2+\frac1{16}.\]Denote the left-hand side by \(c\). We can find points on the hyperbola \(c^2-b^2=\frac1{16}\) by first parametrizing the unit circle and noticing that\[c^2-b^2=\frac1{16}\iff\left(\frac bc\right)^2+\left(\frac1{4c}\right)^2=1.\]The parametrization becomes (we choose \(2t\) rather than \(1-t^2\) in the denominator)\[(b,c)=\left(\frac{1-t^2}{8t},\frac{1+t^2}{8t}\right).\]and thus the completed square can be written (we arbitrarily choose a plus sign when taking roots)\begin{align*}a^2&=\frac{\dfrac{(1-t^2)^2}{16t^2}+1}4+\frac{1+t^2}{8t}=\frac{1+14t^2+t^4+8t+8t^3}{64t^2}\\&=\frac{(1+t)^4+4t(1+t)^2}{64t^2}=\frac{(1+t)^2}{64t^2}((1+t)^2+4t).\end{align*}We now turn to the equation\[d^2=(1+t)^2+4t=(t+3)^2-8=e^2-8.\]In this case let's try a different substitution:\begin{align*}(d,e)&=(u+v,u-v)\\uv&=-2\\(d,e)&=\left(u-\frac2u,u+\frac2u\right)=\left(\frac{u^2-2}u,\frac{u^2+2}u\right)\\t&=e-3=\frac{u^2-3u+2}u=\frac{(u-1)(u-2)}u.\end{align*}Finally,\begin{align*}(a,b)&=\left(\frac{1+t}{8t}d,\frac{1-t^2}{8t}\right)=\left(\frac{\frac{u^2-2u+2}u\cdot\frac{u^2-2}u}{8\frac{(u-1)(u-2)}u},\frac{-\frac{u^2-2u+2}u\cdot\frac{u^2-4u+2}u}{8\frac{(u-1)(u-2)}u}\right)\\&=\frac{u^2-2u+2}{8u(u-1)(u-2)}(u^2-2, -u^2+4u-2).\end{align*}In particular, one possible tetrahedron, obtained when \(u=-1\), has vertices\[[a,b,a^2-b^2]=\frac1{48}[5, 35, -25],\text{ etc}.\]
