$a+1,a-1$ invertible for nilpotent element

Question

Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.

Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.

If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.

But what if $n$ is even?

Answer 1

You know that if $a$ is nilpotent, then $1-a$ is invertible. Hence, also $-(1-a)=a-1$ is invertible. Hence, since $-a$ is also nilpotent, also $1-(-a)=1+a$ is invertible.

Answer 2

If $a$ is nilpotent, then $-a$ is nilpotent. Note that you start from $a^n=0$, not $a^n=1$.