Verified answer

I take it this means

for example ii )

"2 divides greatest common divisor of (n^4) + 1 and (n^2 + 1)"

If 2 divides the greatest common divisor, then it also divides the difference.

Tackling the "2 divides" ones first:

ii)

n^4 + 1 - (n^2 + 1) =

n^4 - n^2 =

n^2 (n^2 - 1)

and since one of those is even

the number is divisible by 2.

iv)

n^3 - 1 - (n^2 + 1) =

n^3 - n^2 - 2 =

n^2 (n - 1) - 2

n^2 is the same parity as n,

and the opposite parity of n-1

so one of n^2 and n-1 is even,

and their product is even,

and product - 2 is even.

v)

n^7 - 1 - n^5 - 1 =

n^7 - n^5 - 2 =

n^5 (n^2 - 1) - 2

and the same argument holds,

since n^5 has the same parity as n,

n^2 has that same parity, and n^2 - 1 has the opposite parity.

Hence the above expression is even.

i)

n^2 + n + 1 =

(n+1)^2 - n =

(n+1)^2 - n - 1 + 1 =

(n+1)^2 - (n+1) + 1

Hence it is always 1 more than a multiple of n+1

and so the greatest common divisor of it with (n+1) = 1.

iii) (5n+4, 6n+1) | 19

There must be an error in this one,

since it fails for n = 1

5n+4 = 9

6n+1 = 7

and gcd(9 , 7) = 1

which 19 does not divide

Similarly:

n ..... 5n+4 .... 6n+1

1 ..... 9 .... 7

2 ..... 14 ... 13

3 ..... 19 .... 19 ... happens to work

4 ..... 24 .... 25

and so on

Usually, 5n+4 and 6n+1

which differ by n - 3

are mutually prime.