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For 7k, (7k)^2 (mod 7)= 0^2 (mod 7)=0, (7k)^3 (mod 7)=0^3 (mod 7)=0
For 7k+1, (7k+1)^2 (mod 7)= 1^2 (mod 7)=1, (7k+1)^3 (mod 7)= 1^3 (mod 7)=1
For 7k+2, (7k+2)^2 (mod 7)= 2^2 (mod 7)=4, (7k+2)^3 (mod 7)= 2^3 (mod 7)=1
.........
For 7k+6, (7k+6)^2 (mod 7)= 6^2 (mod 7)=1, (7k+6)^3 (mod 7)= 6^3 (mod 7)=6
for any 7k+r, if the result is r, then it has the same form.
Well x=n^6...... Now, If n=0 MOD 7 then , n^6=0 MOD 7
Thus, x can be expressed as 7k.... Now if n=1,2,3,4,5,6 MOD 7
Then n^6=1 MOD 7 ---> BUT I HAVEN'T PROVED THIS PART.
let,n = 7k+r (r = 1,2,3,4,5,6)
n^2 = 7x + r^2
n^3 = 7y + r^3
plugin r = 1,2,3,4,5,6 and the intersection of remainders gives the form of a number thats both square and cube
n^p-n strings involving 2 or more colours are partitioned into disjoint sets of p strings each set being strings obtained from each other by by a sequence of cycles ie p divides n^p-n which is Fermat's Little Theorem
You make a string of p beads and join the ends, so that's n^p strings.
Of those, n strings are beads of 1 colour alone (one for each colour)
So n^p-n strings having at least 2 colours etc etc....