can you prove that a^n + b^n is divisible by a+b when n is odd?

when n=1 a^1+b^1 is div by a+b holds

when n=3 a^3+b^3=(a+b)(a^2-ab+b^2) holds

let it be true for some odd no.2k+1

a^(2k+1)+b^(2k+1) is div by a+b

say a^(2k+1)+b^(2k+1)=C(a+b)

a^(2k+1)=C(a+b)-b^(2k+1)...............(1)

To prove it is true for 2k+3

a^(2k+3)+b^(2k+3)

=a^(2k+1)*a^2+b^(2k+1)*b^2

substituting the value of a^(2k+1) from (1)

C(a+b)-b^(2k+1)*a^2+b^(2k+1)*b^2

=C(a+b)-b^(2k+1)(a^2-b^2)

=C(a+b)-b^(2k+1)(a+b)(a-b)

now both the terms are multiples of (a+b)

hence it is true for a(2k+3)+b^(2k+3)

so true for any odd 'n'

Others have proved the same but I can prove it slight differently

let f(a) = a^n + b^n

then f(-b) = (-b)^n + b^n

= -b^n + b^n as n is odd (-1)^n = -1

= 0

by remainder theorem a+b is a factor or the given expression is divisible by a+b when n is odd.

i

By Binomial theorem

a^n + b^n = (a+b)^n + sum(k=1 to k=n-1) [(n_c_k)(a^k)(b^(n-k))]

[Here, n_c_k is n! / (k! * (n-k)!)]

Now, we concentrate on the left most part, sum(k=1 to k=n-1) [(n_c_k)(a^k)(b^(n-k))].

If n is even:

Then, sum(k=1 to k=n-1) [(n_c_k)(a^k)(b^(n-k))] contains odd number of term, this odd number is n-1.

[Argument : If you subtract 1 from any even number it turns to be an odd number]

Also, the fact that [(n_c_k)(a^k)(b^(n-k)]/(a+b) will not give you any series or anwer that you can express in terms or powers of (a+b).

So, now its clear that a^n + b^n is not divisible by a+b when n is even.

If n is odd:

Then, sum(k=1 to k=n-1) [(n_c_k)(a^k)(b^(n-k))] contains even number of term, this odd number is n+2.

[Argument : If you subtract 2 from any even number it will remain even, except for 2.]

Now, the sum will have terms with following conditions

1. There will be 2 terms with same coefficient.

2. The power of a in one term will have equal to power of b in the other term (say p1) and vice versa(say p2).

This allows you express this couple of terms with above property in, product of- common coefficient, a with power p1 -1, b with power p2 -1. And most importantly (a+b).

This (a+b) appearing idicates that the sum, sum(k=1 to k=n-1) [(n_c_k)(a^k)(b^(n-k))] is divisible by (a+b).

Also, trivially (a+b)^n is divisible by (a+b).

Thus, its clear that a^n + b^n is divisible by a+b when n is even.