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
]]>(use mathematical induction)
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