For any odd positive integer ' n ' an+bn is divisible by (a+b). To prove this identity, we can use the factorization: an+bn=(a+b)(an−1−an−2b+an−3b2−...+bn−1) which can be derived using the sum of powers formula for an and bn: an+bn=(a+b)(an−1−an−2b+an−3b2−...+abn−2+bn−1) Since n is odd, the expression inside the parentheses is a sum of an odd number of terms, so it can be written as: an−1−an−2b+an−3b2−...+abn−2+bn−1=(a−b)(an−2+an−4b2+...+bn−2) Now we can substitute this factorization back into the original expression: an+bn=(a+b)(a−b)(an−2+an−4b2+...+bn−2) Since n is odd, a+b is a factor of an+bn by the sum of powers formula. Therefore, we can conclude that an+bn is divisible by a+b for any odd positive integer n.