I have to find a nonzero nilpotent element in the ring $\mathbb{Z}/ 96\mathbb{Z}$.

If we take $a\in \mathbb{Z}/ 96\mathbb{Z}$, $a$ is nilpotent if there exists $n\in\mathbb{Z}$ such that $a^n = 0$.

I have tried with a lot of different elements but I can't find it… Is there any trick to find a nilpotent element in $\mathbb{Z}/ 96\mathbb{Z}$ ?

## Best Answer

Note that $96=32×3=2^5×3$. Note that $6^5=2^5×3^5$. So $6$ is a nulpotent element. In fact any $a$ that is a multiple of both $2$ and $3$ will do, such as $a=6$.

Can you generalize to $\mathbb{Z}/M\mathbb{Z}$; $M$ any integer that is

notsquare-free?