Sophie Germain Identity - Math All Star

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

$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^4+4}$ .