Question Number 108326 by abony1303 last updated on 16/Aug/20
Commented by abony1303 last updated on 16/Aug/20
$$\mathrm{pls}\:\mathrm{help} \\ $$
Answered by mr W last updated on 16/Aug/20
$${a}={pq} \\ $$$${b}>{a}\:\Rightarrow{b}>{pq} \\ $$$$ \\ $$$${a}^{\mathrm{2}} ={p}^{\mathrm{2}} {q}^{\mathrm{2}} ={nb} \\ $$$$\Rightarrow{b}={p}^{\mathrm{2}} {q},\:{n}={q} \\ $$$$\Rightarrow{b}={pq}^{\mathrm{2}} ,\:{n}={p} \\ $$