nt.number theory - b^(n-1)=-1 mod n

It's clear that b = n-1 with n even gives a solution. But there are many other solutions. Here are the solutions $(b,n)$ not of the form $(2k-1, 2k)$, with n less than or equal to 200, from MAPLE.

L := []: for n from 2 to 200 do 
for b from 1 to n-2 do 
if (b^(n-1) mod n) = n-1 then L := [op(L), [b,n]]; fi:
od: od:
L;
[[3, 28], [19, 28], [23, 52], [43, 52], [17, 66], [29, 66], [35, 66],
[41, 66], [19, 70], [59, 70], [27, 76], [31, 76], [31, 112], [47, 112], 
[99, 124], [119, 124], [49, 130], [69, 130], [11, 148], [27, 148], [87, 154], 
[131, 154], [7, 172], [123, 172], [63, 176], [79, 176], [95, 176], [127, 176],
[23, 186], [29, 186], [77, 186], [89, 186], [29, 190], [59, 190], [69, 190], 
[79, 190], [89, 190], [109, 190], [129, 190], [179, 190], [19, 196], [31, 196]]

For example, $3^{28-1} equiv -1 mod 28$, so the pair [3,28] is on the list.

I can't make sense of this output myself, but maybe someone else can?