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**Sophie Germain Identity**

The following daily problem published earlier this week:

(2849) Prove that if $n>1$, then $(n^4 + 4^n)$ is a composite number. |

The most common way to prove an integer is composite is to show that the target has two factors. Therefore, this problem is essentially a polynomial factorization exercise.

When the $4^{th}$ power appears in a factorization problem, the Sophie Germain identity may be relevant. This identity is easy to prove.

(3863) (Sophie Germain's Identity) Prove $a^4 + 4b^4 = (a^2+2b^2-2ab)(a^2+2b^2+2ab)$. |

For example, for problem #2849: when \(n\) is odd, i.e. \(n=2m+1\), the given expression can be factorized using the Sophie Germain identity:

$$n^4+4^n=（2m+1)^4+4^{2m+1}=(2m+1)^4 + 4\times (2^m)^4=\cdots$$

Application of this identity is also seen in solving some competition problems. One typical example appears in 1987 AIME:

(2850) Compute $$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$$ |

Harvard-MIT competition also has one in 2008:

(3271) Evaluate the infinite sum $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^4+4}$. |

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