remainder of the product of the relatively prime numbers

[whodsow]

Hi all, I had a problem, pls help me.

Let b_1 < b_2 < \cdots < b_{\varphi(m)} be the integers between 1 and m that are relatively prime to m (including 1), of course, \varphi(m) is the number of integers between 1 and m that are relatively prime to m, and let B = b_1b_2b_3{\cdots}b_{\varphi(m)} be their product.
Please find a pattern for when B\equiv1 ({\bmod}\ m) and when B\equiv-1 ({\bmod}\ m).

Thanks and Regards.

[CRGreathouse]

ff(m)=centerlift(prod(b=2,m,if(gcd(m,b)==1,b,1),Mo d(1,m)))
for(m=2,20,print(m" "ff(m)))
in Pari/gp gives
2 1
3 -1
4 -1
5 -1
6 -1
7 -1
8 1
9 -1
10 -1
11 -1
12 1
13 -1
14 -1
15 1
16 1
17 -1
18 -1
19 -1
20 1

Does that help?

[whodsow]

Thank you, I've gotten the similar table.

m phi rem
1 1 1
4 2 -1
6 2 -1
8 4 1
9 6 -1
10 4 -1
12 4 1
14 6 -1
15 8 1
16 8 1
18 6 -1
20 8 1
21 12 1
22 10 -1
24 8 1
25 20 -1
26 12 -1
27 18 -1
28 12 1
30 8 1
32 16 1
33 20 1
34 16 -1
35 24 1
36 12 1
38 18 -1
39 24 1
40 16 1
42 12 1
44 20 1
45 24 1
46 22 -1
48 16 1
49 42 -1
50 20 -1
51 32 1
52 24 1
54 18 -1
55 40 1
56 24 1
57 36 1
58 28 -1
60 16 1
62 30 -1
63 36 1
64 32 1
65 48 1
66 20 1
68 32 1
69 44 1
70 24 1
72 24 1
74 36 -1
75 40 1
76 36 1
77 60 1
78 24 1
80 32 1
81 54 -1
82 40 -1
84 24 1
85 64 1
86 42 -1
87 56 1
88 40 1
90 24 1
91 72 1
92 44 1
93 60 1
94 46 -1
95 72 1
96 32 1
98 42 -1
99 60 1

but I still can't find the ppattern, and I have only found that if 4 does not divide \phi(m) then B{\equiv}-1, and if B{\equiv}1, there must be that 4 divides \phi(m)

[whodsow]

Media Man told me it is a generalization of Wilson's Theorem(http://mathworld.wolfram.com/WilsonsTheorem.html)
Thanks.