# remainder of the product of the relatively prime numbers

by whodsow
Tags: congruence, euler's formula
 P: 12 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.
 Sci Advisor HW Helper P: 3,680 ff(m)=centerlift(prod(b=2,m,if(gcd(m,b)==1,b,1),Mod(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?
 P: 12 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 $$\phi(m)$$
P: 12

## remainder of the product of the relatively prime numbers

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

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