c) x^n+y^n isdivisiblebyx+y for oddnaturalnumber n>=1

Answer 1

Let's assume that x+y is divisible by x+y. This means that there exists an integer k such that x+y = k(x^n+y^n).

Rearranging this equation, we get x+y=kx^n+ky^n.

Since n is an odd natural number, we can rewrite this equation as x+y=k(x^n+y^n)=k(x^n+(-y)^n).

Let's consider the sum of the two terms on the right side, x^n+(-y)^n. By the binomial theorem, we can expand this as follows:

x^n+(-y)^n = (x+(-y))(x^(n-1)+(-y)^(n-1)) - xy(x^(n-2)+(-y)^(n-2)) + ... + (-1)^(n-1)xy^(n-2) + (-1)^n y^n.

Notice that each term has a factor of (x+(-y)), so we can rewrite the equation as follows:

x+y = k(x+(-y))A, where A = (x^(n-1)+(-y)^(n-1)) - xy(x^(n-2)+(-y)^(n-2)) + ... + (-1)^(n-1)xy^(n-2) + (-1)^n y^n.

Since x+y = k(x+(-y))A, we can cancel out the factor of (x+(-y)) on both sides of the equation to get 1 = kA.

This implies that k must be 1, which means that x+y = x^n+y^n.

Therefore, x^n+y^n is divisible by x+y for odd natural number n>=1.

Answer 2

You say that
x+y=k(x^n+y^n)=k(x^n+(-y)^n)
but that is not true. It should be
x+y=k(x^n+y^n)=k(x^n-(-y)^n)

Answer 3

You are correct, my mistake. The correct equation should be:

x+y=k(x^n+y^n)=k(x^n-(-y)^n).

Thank you for catching that error. The rest of the proof follows the same steps as described earlier.