Why n⁸ + n⁴ + 1 is always odd, composite (|n|>1) and ends with 1 or 3?

a) n⁸ + n⁴ + 1 = n⁴(n⁴ + 1) + 1

This is an odd number multiplied by an even number (where n is even), plus one, or, an even number multiplied by an odd number (when n is odd), plus one. No matter the order, odd time even is even, so it is an even number plus one, hence it is odd.

b) n⁸ + n⁴ + 1 = (n⁴-n²+1)(n²-n+1)(n²+n+1) = (n²(n-1)(n+1)+1)*(n(n-1)+1)*(n(n+1)+1)

When n=0 this becomes 1*1*1 = 1, and three ones mean this is not composite at all.

When n=1 this becomes 1*1*3 = 3, and two ones mean that this is not a composite factorization.

When n=-1 this becomes 1*3*1 = 3, and two ones again mean that this is also not a composite factorization.

This eliminates all values of n for which at least one of n²(n-1)(n+1), n(n-1), and n(n+1) have a zero value,

and as n⁴>n²>1 where |n|>1, all three terms (n⁴-n²+1),(n²-n+1),and (n²+n+1) are positive and greater than one [Actually, they are at least 13, 3, and 3 respectively, but not all at once.].

Hence n⁸ + n⁴ + 1 = (n⁴-n²+1)(n²-n+1)(n²+n+1) is composite for any integer such that |n| > 1.

c) When n mod 10 = 0, n⁸ mod 10 + n⁴ mod 10 + 1 mod 10 = 0 + 0 + 1 = 1, so the last digit would be 1.

Otherwise, when n mod 5 = 0, n⁸ mod 10 + n⁴ mod 10 + 1 mod 10 = 5 + 5 + 1 = 11, so the last digit would also be 1.

Otherwise, when n mod 10 is even (2,4,6,8), n⁸ mod 10 + n⁴ mod 10 + 1 mod 10 = 6 + 6 + 1 = 13, so the last digit would also be 3.

Otherwise, when n mod 10 is odd (1,3,7,9), n⁸ mod 10 + n⁴ mod 10 + 1 mod 10 = 1 + 1 + 1 = 3, so the last digit would also be 3.

This exhausts all possible unit values of n, so if n mod 5 = 0 then (n⁸ + n⁴ + 1) mod 10 = 1, else (n⁸ + n⁴ + 1) mod 10 = 3, hence n⁸ + n⁴ + 1 always terminates with either a 1 or a 3 as its unit digit.

a) n⁸ and n⁴ are either both even or both odd. That's your proof.

b) don't understand

c) I'm afraid you will just have to list the end digits one by one starting with 0 and all the way to 9. I don't see any other shortcut.

0 --> 0 + 0 + 1

1 --> 1 + 1 + 1

2 --> 6 + 6 + 1

and so on

Recall that n⁸ = (n⁴)², so that should shorten your work.

n⁸ + n⁴ + 1

= (n⁴ + 1)² - n⁴

= (n⁴ + 1)² - (n²)².

a² - b² = (a+b)(a-b).

= (n⁴ + 1 + n²)(n⁴ + 1 - n²).

If |n|>1, then both factors are positive and not 1 making the product composite.