# What is the value of 81^n - 64^n ?

2
by ankitkumar0102

2014-12-17T18:39:53+05:30
81^n - 64^n = 17 k ( where k is any number.)
given below
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.

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