## Fourth power plus four

Let's make some modifications to expression n^{4} + 4, to put it in the form of a product.

We'll start by including the terms + 4n^{2} and -4n^{2}, which does not change the result of the expression, since the sum of both is zero. From there we can apply the factorization rules.

no^{4} + 4 = | no^{4} + 4n^{2} + 4 - 4n^{2} |

= | (n^{2} + 2)^{2} - (2n)^{2} difference of two squares |

= | ((n^{2} + 2) + 2n) ((n^{2} + 2) - 2n) |

= | (n^{2} + 2n + 2) (n^{2} - 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 n^{2} + 2n + 2> 1, for n greater than or equal to 1.

Therefore, we must have the second factor equal to 1, ie:

no^{2} - 2n + 2 = 1

no^{2} - 2n + 1 = 0 (subtracting 1 on both sides)

(n - 1)^{2} = 0 => n = 1

When n = 1, we have a prime number, because n^{4} + 4 = 5. Therefore, this is the only value of n for which n^{4} + 4 is cousin.

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