Question

Show that 2²⁴-1 and 2¹⁶-1 are composite. Hints:use expansion of (a²-b²)

Answer 1

(1) $\left(2^{24}-1\right)$

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

$\begin{array}{rl} \therefore 2^{24}-1=\left(2^{12}\right)^{2}-1^{2}=\left(2^{12}-1\right)\left(2^{12}+1\right) \\ =\left(\left(2^{6}\right)^{2}-1^{2}\right)\left(2^{12}+1\right) \\ =\left(2^{6}-1\right)\left(2^{6}+1\right)\left(2^{12}+1\right) \\ =\left(2^{3}-1\right)\left(2^{3}+1\right)\left(2^{6}+1\right)\left(2^{12}+1\right) \\ = 7 \times 9 \times\left(2^{6}+1\right)\left(2^{12}+1\right) \end{array}$

$\therefore 2^{24}-1$ has the divisor $7,9,2^{6}+1,2^{12}+1$ other than the $1 \& 2^{24}-1$ .

$\therefore 2^{24}-1$ is composite number.

(2) $2^{16}-1$

$\begin{aligned} 2^{16}&-1=\left(2^{8}\right)^{2}-1^{2}=\left(2^{8}-1\right)\left(2^{8}+1\right) \\ &=\left(2^{4}-1\right)\left(2^{4}+1\right)\left(2^{8}+1\right) \\ &=\left(2^{2}-1\right)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right) \\ &=3 \times 5 \times 17 \times\left(2^{8}+1\right) \end{aligned}$

 

$\therefore 2^{16}-1$ has the divisor other than 1 & $2^{16}-1$ .

$\therefore 2^{16}-1$ is composite number.