Fourth power plus four
Let's make some modifications to expression n4 + 4, to put it in the form of a product.
We'll start by including the terms + 4n2 and -4n2, which does not change the result of the expression, since the sum of both is zero. From there we can apply the factorization rules.
|no4 + 4 =||no4 + 4n2 + 4 - 4n2|
|=||(n2 + 2)2 - (2n)2 difference of two squares|
|=||((n2 + 2) + 2n) ((n2 + 2) - 2n)|
|=||(n2 + 2n + 2) (n2 - 2n + 2)|
Then we come to a product, which can only result in a prime number if one of the factors is equal to 1.
The first factor does not satisfy this condition, since we clearly realize that n2 + 2n + 2> 1, for n greater than or equal to 1.
Therefore, we must have the second factor equal to 1, ie:
no2 - 2n + 2 = 1
no2 - 2n + 1 = 0 (subtracting 1 on both sides)
(n - 1)2 = 0 => n = 1
When n = 1, we have a prime number, because n4 + 4 = 5. Therefore, this is the only value of n for which n4 + 4 is cousin.
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