Let : $A=\frac{2^{4n+2}+1}{5}$ , $n>1$

Question

Prove that the number A is not primary

Such that :

$A=\frac{2^{4n+2}+1}{5}$

$n≥2$

n=2 then $A=205$

Please I need some ideas to approach it

Answer 1

The somewhat well-known identity $4n^4+1 =(2 n^2 - 2 n + 1) (2 n^2 + 2 n + 1) $ gives

$\begin{array}\\ 2^{4n+2}+1 &=4(2^n)^4+1\\ &=(2\cdot 2^{2n}-2\cdot 2^n+1)(2\cdot 2^{2n}+2\cdot 2^n+1)\\ &=(2^{2n+1}-2^{n+1}+1)(2^{2n+1}+2^{n+1}+1)\\ \end{array} $

and both factors are greater than 5 for $n \ge 2$.

Answer 2

First, the numerator is divisible by $5$. This follows since $4^{2n+1}+1=(4+1)(4^{2n}-4^{2n-1}+\dots-4+1)$.

Second, @marty cohen has shown that the numerator is the product of factors each greater than five.

It follows that $A\in\Bbb N$ is not prime.