Given $ x^{2}+px+q = 0 $ with discriminant $v=\sqrt{p^{2}-4q}$ and solutions $x_{1}=\frac{-p+v}{2},x_{2}=\frac{-p-v}{2}$. Given $ y^{2}-py+r = 0 $ with discriminant $w=\sqrt{p^{2}-4r}$ and solutions $y_{1}=\frac{p+w}{2},y_{2}=\frac{p-w}{2}$. Given $ x_{1}y_{1}-x_{2}y_{2}= 1 $, so: \[\frac{-p+v}{2} \cdot \frac{p+w}{2} - \frac{-p-v}{2} \cdot \frac{p-w}{2} = 1 \] Calculating: $pv=pw+2$; squaring: \[p^{2}(p^{2}-4q)=p^{2}(p^{2}-4r)+4+4p\sqrt{p^{2}-4r}\] \[p^{2}(r-q)-1=p\sqrt{p^{2}-4r}\] \[p^{4}(r-q)^{2}-2p^{2}(r-q)+1=p^{2}(p^{2}-4r)\] \[p^{4}(r-q)^{2}+2p^{2}(r+q)+1 = p^{4} \]