for what value(s) of $x$ is $nx$ congruent to $1 \pmod {(n+1)}$

Question

I need to find some fixed integer value for $x$ which satisfies $ nx \equiv 1 \pmod{ n+1} $. This is for a midterm review and I dont really see how this is possible without using $n$ in the formula, which is forbidden. Any help? thanks

Answer 1

$n+1\equiv 0 \ ( \textrm{mod } (n+1) )$

$nx+x\equiv 0 \ ( \textrm{mod } (n+1) )$

$nx\equiv -x \ ( \textrm{mod } (n+1) )$

So you need that $-x\equiv 1 \ ( \textrm{mod } (n+1) )$, which means $x\equiv -1 \ ( \textrm{mod } (n+1) )$. In conclusion, the values of $x$ you want are $\{(n+1)k-1: \ k\in\mathbb{Z} \}$.