b^(n-1)=-1 mod n By Fermat's little theorem we know that
$$b^{p-1}=1 \mod p$$
if p is prime and $\gcd(b,p)=1$.  On the other hand, I was wondering whether
$$b^{n-1}=-1 \mod n$$
can occur at all?
Update: sorry, I meant n odd.  Please excuse.
 A: There are no solutions to $b^{n-1}\equiv-1\pmod n$ with $n$ odd. 
Let $n>1$ be odd. Every prime dividing $n$ can be written as $2^km+1$ for some positive $k$ and some odd integer $m$. Among those primes, let $p$ have the minimal value of $k$. Then $n-1=2^kr$ for some integer $r$. If 
$b^{n-1}\equiv-1\pmod n$ then $b^{n-1}\equiv-1\pmod p$ so $b^{(n-1)m}\equiv(-1)^m\equiv-1\pmod p$ and 
$\gcd(b,p)=1$. But $b^{(n-1)m}=b^{2^kmr}=b^{(p-1)r}\equiv1\pmod p$ by little Fermat. Contradiction, QED. 
A: That would be equivalent to $2(n-1) = k\varphi(n)$ and $n-1\ne k'\varphi(n)$ by Fermat's little theorem for composite numbers. 
The second condition is equivalent to being able to satisfy first with $k$ odd, so we could try $k = 3$. Thus we have $n = 3n' +1$ and $2n' = \varphi(3n' + 1)$. Now the trivial choice for $n' =1$ works!
Thus we find $n = 4$: $$(-1)^{(4-1)} = -1 (\mathop{\text{mod}} 4).$$
A: This is something that brute force can answer:

In[1]:= Reap[Do[
   If[Mod[PowerMod[a, n - 1, n] + 1 , n] == 0, Sow[{a, n}]],
   {n, 2, 20},
   {a, 1, n - 1}
   ]][[2, 1]]

Out[1]= {{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}, {13, 14}, 
         {15, 16}, {17, 18}, {19, 20}}

(This lists the pairs $(b,n)$ for $n$ at most 20)
