The Sophie Germain Identity

Feb 05, 2016

Marie Sophie Germain was a nineteenth century mathematician who did extensive work on the theory of elasticity and on Fermat's Last Theorem.

Let’s look at this question which was first appeared in the Mathematics Magazine 1950. It was put forward by A.Makowski, a leader of the Polish IMO team. Later it was asked for the 1978 Ku:rschak competition.

Prove that is composite for all natural numbers $n>1$.

This question is a particular example of the Sophie Germain’s Identity which states that, \begin{align} a^4 + 4b^4 = (a^2 + 2b^2 -2ab)(a^2 + 2b^2 + 2ab). \end{align}

Marie Sophie Germain was an nineteenth century mathematician who did extensive work on the theory of elasticity and on the Fermat’s Last Theorem. Her work enabled [TODO:find name] to prove Fermat’s Last Theorem for $n= 5$. For more information about Sophie Germain go this excellent Mactutor biography[TODO:link]. So let’s use the Sohpie Germain Identity to solve this question.

##Proof

Consider the special case when $n$ is even. Then,

\begin{align} 2 \mid (n^4 + 4^n). \end{align}

But what if $n$ is odd? Let $n = 2k+1$. Let’s apply the Sohpie Germain Identity and see what we get. Notice that $n^4 + 4^n$ can be rewritten as, \begin{align} (2k+1)^4 + 4\dot (2^k)^4. \end{align} Which means that we can factorize the expression into,

\begin{equation} (2k+1)^2 + 2(2^k)^2 - 2(2k+1)(2^k) \times \\ (2k+1)^2 + 2(2^k)^2 + 2(2k+1)(2^k) \end{equation}

So we realize that the expression can indeed be factorized.

It’s not too hard too figure out why $n$ cannot be $1$.