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Topic review (newest first)
- ganesh
- 2005-09-16 00:04:44
ganesh wrote:
Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= {[k²(k+2)+k(k+2)]+k+1}mod6
= {[(k²+k)(k+2)] +k+1}mod6
={[(k(k+1)(k+2) + k+1}mod6
We know that k(k+1)(k+2) is divisble by 6 for any k>1, k∈N,
Hence the above is reduced to
(k+1)mod6
It is seen that it is true for k+1, hence, it is true for any value of k.
q.e.d 
- John E. Franklin
- 2005-09-15 14:43:41
P(n+1) is (n^3 + 3n^2 + 3n + 1)%6 = (n+1)%6
and assume we start with n^3%6 = n%6, where % means mod (like in C language).
(n^3 + 3n^2 + 2n)%6 + (n+1)%6 = (n+1)%6
(n^3 + 3n^2 + 2n)%6 = 0
Substitute n%6 in place of n^3%6 and get:
(3n(n+1))%6=0
For even numbers, 3n is a multiple of 6, so that works.
For odd numbers, the (n+1) part is even, so that works.
- ganesh
- 2005-09-15 14:16:20
Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= [k(k^2+2)+k(k+2)+k+1]mod6...
Running out of time...gotta leave....
- daisy
- 2005-09-15 12:59:37
prove that for every non-negative integer n , n^3 mod 6 = n mod 6.
(use mathematical induction)