This congruence may generate the trivial n = 1 ⋅ n {\displaystyle \textstyle n=1\cdot n} ; in this case we try to find another suitable congruence.
Denna kongruens kan trivialt generera n = 1 ⋅ n {\displaystyle \textstyle n=1\cdot n} ; i det här fallet försöker vi hitta en annan lämplig delmängd.