Question

If a and b are distinct integers, prove that a – b is a factor of a n – b n , whenever n is a positive integer. [ Hint: write a n = ( a – b + b ) n and expand]

Answer 1

The given expression is ( a n − b n ) , and we have to prove that ( a−b ) is a factor of it where n is positive integer.

a n can be written as, a n = ( a−b+b ) n

a n = ( a−b+b ) n = [ ( a−b )+b ] n = C n 0 ( a−b ) n + C n 1 ( a−b ) n−1 b+.......+ C n n−1 ( a−b ) b n−1 + C n n b n = ( a−b ) n + C n 1 ( a−b ) n−1 b+.......+ C n n−1 ( a−b ) b n−1 + b n

Taking bnto L.H.S, and taking (a – b) common from R.H.S, we get

( a n − b n )=( a−b )[ ( a−b ) n−1 + C n 1 ( a−b ) n−2 b+........+ C n n−1 b n−1 ] ( a n − b n )=k( a−b )

Where, k=[ ( a−b ) n−1 + C n 1 ( a−b ) n−2 b+........+ C n n−1 b n−1 ] is a natural number.

Hence, it is proved that the expression ( a−b ) is factor of ( a n − b n ) , where n is a natural number.