# Sophie Germain's Identity

An identity named after the famous french mathematician Sophie Germain (1776 - 1831) is a useful tool for math olympians:

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

The proof is by direct verification.

Here's a problem that is easily solved with Sophie Germain's identity:

Is $2015^{4}+4^{2015}$ prime?

Due to Sophie Germain's identity the answer is negative. Indeed,

$\begin{align} 2015^{4}+4^{2015}&= 2015^{4}+4\cdot 4^{2014}\\ &= 2015^{4}+4\cdot 2^{4028}\\ &= 2015^{4}+4\cdot \bigg(2^{1007}\bigg)^{4}\\ &= (2015^{2}+2^{2015}- 2015\cdot 2^{1008})(2015^{2}+2^{2015}+ 2015\cdot 2^{1008}). \end{align}$

Since the second factor is clearly greater than $1,$ we only need to check that this is also true for the first factor. But $2^{2007}\gt 2^{11}=2048\gt 2015,$ implying that $2^{2015}\gt 2015\cdot 2^{1008}.$

In fact, $n^{4}+4^{n},$ where $n$ is a positive integer, is never a prime. It is even if $n$ is even, and, if $n$ is odd, we can simply generalize the above derivation.

### References

- A. Engel,
*Problem-Solving Strategies*, Springer Verlag, 1998, p 121

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

65462498 |