Question

[KAl(SO4)2·12H2O]

Tìm \(n\in N\)để \(5^{2n^2-6n+2}-12\)là số nguyên tố.

Answer 1

[minhduc]

Đặt A = 52n2−6n+2−12=25n2−3n+1−12≡12n2−3n+1−12(mod13)52n2−6n+2−12=25n2−3n+1−12≡12n2−3n+1−12(mod13)

                    =>12n2−3n+1−12=12.(12n(n−3)−1)12n2−3n+1−12=12.(12n(n−3)−1)

                   (12n(n−3)−1)(12n(n−3)−1) chia luôn chia 13 dư 1 do n(n-3) luôn chia hết cho 2

                   => 52n2−6n+2−12⋮1352n2−6n+2−12⋮13 mà A lại là số nguyên tố nên A= 13 

                  =>  52n2−6n+2=2552n2−6n+2=25 => n =3

               Vậy n = 3

Answer 2

[minhduc]

$n^2-3n+1=n^2-n-2n+1$ là số lẻ nên ta có $5^{2n^2-6n+2}-12\equiv 1-1^{n^2-3n+1}\equiv 0\left(\mathrm{mod}13\right)$

Do đó $5^{2n^2-6n+2}-12=13\iff 5^{2n^2-6n+2}=25\iff 2n^2-6n+2=2\iff n=0$ hoặc $n=3$