Let n=1 then, the value of xn+yn becomes, x1+y1=x+y It is divisible by x+y. Let n=2, its value will be x2+y2 which is not divisible by x+y Let n=3 then, the value of xn+yn will be, x3+y3=(x+y)(x2−xy+y2) Which is divisible by x+y. Thus, xn+yn is divisible by x+y only when n= odd This, implies option (2) and (4) are wrong. The xn+yn is not divisible x−y for all n∈N. Therefore, the correct option is (3).