Those who love (elementary) problem-solving eventually come across the Sophie Germain identity. It has lots of applications in elementary number theory, algebra and so on. The identity states

$x^4 + 4y^4 = (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$.

Indeed, note that

$x^4 + 4y^4$

$= (x^2)^2 + (2y^2)^2 + 2\cdot x^2 \cdot 2y^2 - 2\cdot x^2 \cdot 2y^2$

$= (x^2 + 2y^2)^2 - (2xy)^2$

$= (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$

More can read about Marie-Sophie Germain, a brilliant mathematician, here and here.

Let us use the above identity in solving a couple of problems. Here is the first one.

Problem 1: Evaluate $\displaystyle \sum_{k=1}^{n}\frac{4k}{4k^4 + 1}$.

Solution: A first glance tells us that the sum should somehow “telescope.” But the denominator looks somewhat nasty! And, it is here that the above identity comes to our rescue. Using the Sophie Germain identity, we first note that

$1 + 4k^4 = (1 -2k + 2k^2)(1 + 2k + 2k^2).$

We thus have

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

$\displaystyle = \sum_{k=1}^{n} \left(\frac{(1 + 2k + 2k^2)-(1 -2k + 2k^2)}{(1 -2k + 2k^2)(1 + 2k + 2k^2)}\right)$

$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 + 2k + 2k^2}\right)$

$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 - 2(k+1) + 2(k+1)^2} \right)$

$\displaystyle = 1 - \frac1{1 + 2n + 2n^2}$.

Here is another one.

Problem 2: Show that $n^4 + 4^n$ is a prime iff $n=1$, where $n \in \mathbb{N}$.

Solution: Note that if $n$ is even, then the expression is clearly composite. If $n$ is odd, say, $2k+1$ for some $k \in \mathbb{N}$, then we have

$n^4 + 4^n$

$= n^4 + 4^{2k+1}$

$= n^4 + 4\cdot (2^k)^4$

$= (n^2 - n\cdot 2^{k+1} + 2^{2k+1})(n^2 + n\cdot 2^{k+1} + 2^{2k+1})$

And, if $n > 1$, then both the factors above are greater than $1$, and hence the expression is composite. Moreover, if $n = 1$, we have $1^4 + 4^n = 5$, which is prime. And, we are done.