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2014-12-17T18:39:53+05:30
81^n - 64^n = 17 k ( where k is any number.)
1 3 1
show me the process please
given below
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2014-12-18T01:19:08+05:30

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N = 81^n - 64^n ,   let us say n is a non-negative integer,  0,1,2,3....

N = (81-64) * (81^{n-1}+81^{n-2}*64+81^{n-3}*64^2+....+81^2*64^{n-3}+81*64^{n-2}+64^{n-1})

N = 9^{2n} - 8^{2n}
    = (9^n)2 - (8^n)^2
    = (9^n + 8^n)  (9^n - 8^n)
    = (3^{2n} + 2^{3n}) [ 3^{2n} - 2^{3n} ]        there are minimum two factors.

n = 0,  N = 1 - 1 = 0
  = 1,  N = 81 - 64 = 17
  = 2 ,  N = 81^2 - 64^2  = (81+64) * (81 - 64)  = 29 * 5 * 17
  = 3,  N =  81^3 - 64^3  =  (81 - 64) * (81^2 + 81 * 64 + 64^2)  = 17 * 73 * 7 * 31
  = 4,   N = 81^4 - 64^4 = (81-64)(81+64)(81^2+64^2) = 17 * 29 * 5 * 10657
n = 5,  N = 17 * 11 * 41 * 491 * 641

n=6,   N = (81^3+64^3)(81-64)(81^2+81*64+64^2) = 17*5473*5*29*73* 7 * 31 
n=7,   N = 17 * 1086985055521 

Always,  A^n - B^n  will have (A-B) as a factor.   Hence the given expression has 17 as a factor.  Perhaps there are other repeating factors for some values of n.

Also it seems to be multiple of odd integers. So it is always an odd integer.

2 5 2
It also seems that the units digit of the result is either 5 or 7.
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