計算1
不失一般性，令\(p<q\)，\(p-1|3pq-1\)，\(p-1|3pq-1-p+1=p(3q-1)\)
由\((p,p-1)=1\)知\(p-1|3q-1\)，同理，\(q-1|3p-1\) (考試的時候只有得到類似的結論就猜11跟17)
\(\displaystyle\frac{3p-1}{q-1}<\frac{3q-1}{q-1}=3+\frac{2}{q-1}\le3+\frac{2}{7-1}=3.\cdots\)
(1) 若\(\displaystyle\frac{3p-1}{q-1}=1\)，\(3p=q\)，不合
(2) 若\(\displaystyle\frac{3p-1}{q-1}=2\)，\(\displaystyle\frac{3q-1}{p-1}=\frac{9q-3}{2q-4}\)
由\(\displaystyle\lim_{q\to\infty}\frac{9q-3}{2q-4}=4.5\)和\(q=7\)時\(\displaystyle\frac{9q-3}{2q-4}=6\)知\(\displaystyle5\le\frac{9q-3}{2q-4}\le6\)
討論可知\(\displaystyle\frac{9q-3}{2q-4}=5\)時\(p=11,q=17\)
(3) 若\(\displaystyle\frac{3p-1}{q-1}=3\)，\(3p=3q-2\)，不合
