回復 1# shmilyho 的帖子
用歸謬證法，先設\({{p}^{2}}\ne 4q\)
判別式
\(\begin{align}
& \left( 1+{{q}^{2}} \right){{p}^{2}}-4\left( 1-q+\frac{{{p}^{2}}}{2} \right)\left[ \left( q-1 \right)q+\frac{{{p}^{2}}}{2} \right] \\
& \ne 4q\left( 1+{{q}^{2}} \right)-4\left( 1-q+\frac{4q}{2} \right)\left[ \left( q-1 \right)q+\frac{4q}{2} \right] \\
\end{align}\)
而\(4q\left( 1+{{q}^{2}} \right)-4\left( 1-q+\frac{4q}{2} \right)\left[ \left( q-1 \right)q+\frac{4q}{2} \right]=0\)
