n is an integer. if n³+n²+n¹+1 is always an odd number, is (n+1)(ⁿ+¹) always an odd number?

Question

n is an integer.

if n³+n²+n¹+1 is always an odd number, is (n+1)(ⁿ+¹) always an odd number?
note:

(n+1) is the power of (n+1), might seem confusing due to my lack of knowledge on how to make a + on power size, or () on PC.

Answer 1

If "n" is an integer: { ...-3,-2,-1,0,1,2,3, ...} it is within this set.

 

So let's start by factoring to determine what rules there are for "n" given n³+n²+n¹+1 is always odd.

 

Note odd/even products and sums: (think of simple examples)

even*odd=even ; even*even=even ; odd*odd=odd

even+even=even ; even+odd=odd ; odd+odd=even

 

n³+n¹+n²+1

n(n²+1)+n²+1

 

if n is odd n² is odd and (n²+1) = even.

if n is even n² is even and (n²+1) = odd. 

so this becomes;

n = odd ; odd(even)+even = even+even = even 

n = even ; even(odd)+odd = even+odd = odd

 

So n must be an even integer for the first statement to be true.

 

(n+1)⁽ⁿ⁺¹⁾ = (odd)^(odd) = always odd