Question

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

Answer 1

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

n2−3n+1=n2−n−2n+1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n2−3n+1=n2−n−2n+1 là số lẻ nên ta có 52n2−6n+2−12=13⇔52n2−6n+2=25⇔2n2−6n+2=2⇔n=0" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">52n2−6n+2−12=13⇔52n2−6n+2=25⇔2n2−6n+2=2⇔n=0 hoặc n=3