Sophie Germain Identity: Factoring a^4+4b^4

Sophie Germain Identity

One of the cleverest, most useful identities I've seen is the Sophie Germain Identity.

Problem: Factor $a^4+4b^4$.

Solution: Factoring these simple expressions has always seen bizarre to me. The trick here is to create a difference of squares. Adding and subtracting $4a^2b^2$

$$a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=(a^2+2b^2)^2-(2ab)^2$$

So

$$a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$$

Problem (BMO 2014): Calculate the value of

$$\frac{2014^4+4\cdot2013^4}{2013^2+4027^2}-\frac{2012^4+4\cdot2013^4}{2013^2+4025^2}$$

Solution: Let $x=2013$. Then we want to find

$$\frac{(x+1)^4+4x^4}{x^2+(2x+1)^2}-\frac{(x-1)^4+4x^4}{x^2+(2x-1)^2}$$

The numerator of the first is

$$\begin{align*} (x+1)^4+4x^4&=((x+1)^2+2(x+1)x+2x^2)((x+1)^2-2(x+1)x+2x^2)\\&=(5x^2+4x+1)(x^2+1)=(x^2+1)(x^2+(2x+1)^2) \end{align*}$$

The numerator of the second is

$$\begin{align*} (x-1)^4+4x^4&=((x-1)^2+2(x-1)x+2x^2)((x-1)^2-2(x-1)x+2x^2)\\&=(5x^2-4x+1)(x^2+1)=(x^2+1)(x^2+(2x-1)^2) \end{align*}$$

So the expression is simply $(x^2+1)-(x^2+1)=0$.

Problem (AIME 1987): Compute

$$\prod_{i=1}^5\frac{(12i-2)^4+324}{(12i-8)^4+324}$$

Solution:

$$x^4+324=x^4+4\cdot3^4=(x(x-6)+18)(x(x+6)+18)$$

so

$$\begin{align} \frac{(x+6)^4+324}{x^4+324}=\frac{((x+6)x+18)((x+6)(x+12)+18)}{(x(x-6)+18)(x(x+6)+18)}&= \\ \frac{(x+6)(x+12)+18}{x(x-6)+18}. \end{align}$$

From here we get

$$\frac{(4+6)(4+12)+18}{4(4-6)+18}\frac{(16+6)(16+12)+18}{16(16-6)+18}\dots\frac{(52+6)(52+12)+18}{52(52-6)+18}$$

which telescopes as

$$\frac{58(64)+18}{4(-2)+18}=373$$

Follow-up: do any other cool identities like Sophie Germain exist? It is derived from $x^2-y^2=((x+y)+\sqrt{2xy})((x+y)-\sqrt{2xy})$ and then substituting to get integer coefficients. I don't think so; I think it is a consequence of difference of squares which is a very fundamental factorization with not many other identities like it.

Sophie Germain Identity is a very common and useful tactic in number theory. It is basically just a factorization identity. \begin{eqnarray*} a^4+4b^4 & = & (a^2+2ab+2b^2)(a^2-2ab+2b^2) \end{eqnarray*} The proof is quite straightforward. It seems to be a fairly regular identity, but it has quite a lot of applications. Let's start with following problem from IMO.

Problem If $n>1$, prove that $n^4+4^n$ is composite.

If $n$ is even, then clearly $n^4+4^n$ is even and greater than $2$, so composite. Let's deal with odd $n=2k+1$ now. \begin{eqnarray*} n^4+4^n & = & n^4+4^{2k+1}\newline & = & n^4+4\cdot(2^k)^4\newline & = & (n^2+2n\cdot2^k+2\cdot(2^k)^2)(n^2-2n\cdot2^k+2(2^k)^2) \end{eqnarray*} $n^2-2n\cdot2^k+2(2^k)^2=(n-2^k)^2+2^{2k}$, this is the smaller part and clearly greater than $1$. Therefore, $n^4+4^n$ is a product of two numbers greater than $1$ and is not a prime.

The following problem is from Russian Olympiad for junior students, which only a few participants could solve.

Problem Is $545^{4}+4^{545}$ a prime?

$545^{4}+4^{545}=\left(545\right)^4+4\cdot\left(4^{136}\right)^4.$ With this I encourage you to use the Sophie Germain Identity to finish the problem.

We are going to derive the Sophie Germain Identity. It seems quite easy to multiply the factorization after you have been given the factorization, but how does one find it in the first place?

When you see $a^4+4b^4$, your first thought should be, "It looks like a perfect square!". Whenever you have just even powers, this should be your first thought. However, it isn't quite a square.

Thus, $a^4+4b^4=a^4+4b^4+4a^2b^2-4a^2b^2$. Look carefully. This is a difference of squares!

Therefore, the entire expression factorizes as $$a^4+4b^4=(a^2+2b^2)^2-(2ab)^2=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$$ which was to be desired.

Remember, look for even powers; these can be signs that an expression is a perfect square. Also, sometimes you must complicate an expression to make it simpler in the end.