Let $n \in \Bbb N$. Find the inverse of $n \pmod {n + 1}$

Question

Let $n \in \Bbb N$. Find the inverse of $n \pmod {n + 1}$

I tried answering the question and got $n+1 \pmod 1$, is this correct? Do I need to use Pell's equation?

Answer 1

We have $-1\cdot n+ 1\cdot (n+1)=1$, which means modulo $n+1$ that $-1\cdot n\equiv 1\bmod n+1$. Hence $-1$ is the inverse.

Answer 2

We have $$n^{-1} \pmod{n+1} \equiv (-1)^{-1}\pmod{n+1} \equiv (-1)\pmod{n+1} = n \pmod{n+1}$$ Also, note that $$n \cdot n = (n+1)(n-1) + 1$$

Answer 3

There is only one inverse for n: It is n itself. For n=-1 mod n+1. And so (-1)(-1)=1 mod n+1