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Forum: Conjectures 'R Us 2009-01-12, 06:22
Replies: 73
Views: 10,242
Posted By S485122
1 == 8 == -6 == 15 == -13 mod 7 is the same as...

1 == 8 == -6 == 15 == -13 mod 7 is the same as saying 1, 8, -6, 15 and -13 are multiples of 7 plus 1.

If we compute (7+1)^m = 7^m+m*7^(m-1)*1+ ... +m*7*1^(m-1)+1^m all terms except 1^m contain a...
Forum: Conjectures 'R Us 2009-01-11, 17:31
Replies: 73
Views: 10,242
Posted By S485122
2^(3*m)=(2^3)^m (2^3)^m=8^m 8^m=(7+1)^m ...

2^(3*m)=(2^3)^m
(2^3)^m=8^m
8^m=(7+1)^m
(7+1)^m==(0+1)^m mod 7
1^m=1
so
2^(3*m)==1 mod 7

Jacob
Forum: Conjectures 'R Us 2009-01-10, 14:13
Replies: 73
Views: 10,242
Posted By S485122
n==1 mod 3. n=3*m+1. ...

n==1 mod 3.
n=3*m+1.
123*2^n-1=123*2^(3*m+1)-1=123*2*2^(3*m)-1
2^3==1 mod 7 => 2^(3*m)=(2^3)^m==1 mod 7
123*2*2^(3*m)-1==246*1-1 mod 7
245==0 mod 7

Use the same path for the other case.
...
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