Some Algebraic Identities for use in number theory — RMO | For the love of Maths: the Queen of all Sciences

Many problems about divisibility can be solved using algebraic identities such as:

{} $a-b|a^{n}-b^{n}$ for all $n \in N$ and $a+b|a^{n}+b^{n}$ for all odd $n \in N$ .
Note that $1+a+\ldots + a^{n}=\frac{a^{n+1}-1}{a-1}$ provided that $n \neq 1$ .
(Sophie Germain’s identity)

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

Here is a problem which can be solved using algebraic identities:

Example:

Show that $n^{5}+n^{4}+1$ is not a prime for $n>1$ .

Solution:

Note that

$n^{5}+n^{4}+1=n^{5}+n^{4}+n^{3}+n^{2}+n+1-(n^{3}+n^{2}+n)$

which in turn equals

$\frac{n^{6}-1}{n-1} - n \frac{n^{3}-1}{n-1}=(n^{3}+1)\frac{n^{3}-1}{n-1}-n\frac{n^{3}-1}{n-1}$ , that is,

$=\frac{(n^{3}-1)(n^{3}-n+1)}{n-1} = (n^{2}+n+1)(n^{3}-n+1)$

Since $n>1$ , each $n^{2}+n+1$ , $n^{3}-n+1$ is greater than 1. Hence, $n^{5}+n^{4}+1$ is not prime for $n>1$ .

More later,

Nalin Pithwa

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google+ account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s