x^(n)+y^(n) is divisible by x+y for odd natural numer n>_1
consider
x^(n-1) - x^(n-2) y + ... + x y^(n-2) - y^(n-1)
Now multiply that by x+y and you have
x*n - x^(n-1) y + ... - x^2 y^(n-2) + xy^(n-1)
+x^(n-1) y - x^(n-2) y^2 + ... + x^2 y^(n-2) + y^n
= x^n + y^n
this is just an extension of sum of cubes, 5th, etc.
x^3 + y^3 = (x+y)(x^2-xy+y^2)
x^5 + y^5 = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
...
Interesting
yes
To prove that x^(n) + y^(n) is divisible by x+y for odd natural numbers n greater than or equal to 1, we can use mathematical induction.
First, let's check the base case when n = 1:
When n = 1, we have x^(1) + y^(1) = x + y, which is clearly divisible by x+y.
Now let's assume that x^(n) + y^(n) is divisible by x+y for some odd value of n, and we want to prove that it is true for n+2 as well.
We start with the expression x^(n+2) + y^(n+2):
x^(n+2) + y^(n+2) = (x^(n) * x^2) + (y^(n) * y^2)
= (x * x^n * x) + (y * y^n * y)
= (x^(n+1) * x) + (y^(n+1) * y)
= x(x^n + y^n) + y(x^n + y^n)
Now, we need to show that x(x^n + y^n) + y(x^n + y^n) is divisible by x + y.
We can factor out (x^n + y^n) from both terms:
x(x^n + y^n) + y(x^n + y^n) = (x + y)(x^n + y^n)
Since we assumed that x^n + y^n is divisible by x + y, we can rewrite it as:
(x + y)(k), where k is some integer.
Therefore, we have shown that x^(n+2) + y^(n+2) is also divisible by x + y.
By using mathematical induction, we have proven that x^(n) + y^(n) is divisible by x + y for all odd natural numbers n greater than or equal to 1.