a

Ta có:\(2020\equiv1\left(mod3\right)\Rightarrow2020^{2019}\equiv1\left(mod3\right)\Rightarrow2020^{2019}-1\equiv0\left(mod3\right)\)

Khi đó:\(\left(2020^{2019}+1\right)\cdot\left(2020^{2019}-1\right)\equiv0\left(mod3\right)\)

suy ra đpcm

b

\(n^5+96n=n\left(n^4+96\right)\)

Để \(n^5+96n\) là số nguyên tố thì:\(n^4+96=1\left(h\right)n=1\)

Do \(n^4+96>1\Rightarrow n=1\)

Thay vào ta thấy thỏa mãn

Vậy n=1

a, =2020^4038 -1

Vì  \(2020 \equiv 1 \pmod{3}\)

->\(2020^(4038) \equiv 1 \pmod{3}\)

->2020^4038 -1 chia hết cho 3 -> dpcm

(2020^2019+1)(2020^2019-1)=(2020^2019+1).(2020-1).(2020^2018 + 2020^2017+ 2020^2016+....+1) 

mà 2019 chia hết cho 3 nên (2020^2019+1).(2020-1).(2020^2018 + 2020^2017+ 2020^2016+....+1) chia hết cho 3

b) n^5 + 96n=n(n^4 + 96) luôn chia hết cho n và (n^4 + 96)

n(n^4 + 96) là số nguyên tố <=> n=1

a) \(\left(2020^{2019}+1\right)\left(2020^{2019}-1\right)=\left(2020^{2019}\right)^2-1=2020^{4038}-1\)

Ta có: 2020 = 1 mod 3

\(\Rightarrow2020^{2019}\equiv1mod3\)

\(\Rightarrow2020^{4038}-1\equiv0mod3\)

=> đpcm

1: \(A=6^{2020}\left(1+6\right)+6^{2022}\left(1+6\right)\)

\(=7\left(6^{2020}+6^{2022}\right)⋮7\)

Bài 1:

$A=6^{2020}(1+6+6^2+6^3)=6^{2020}.259=6^{2020}.7.37\vdots 7$

Ta có đpcm.

Bài 2:

$1+2+3+...+n=1275$

$\frac{n(n+1)}{2}=1275$

$n(n+1)=2.1275=2550$

$n(n+1)=50.51$

$\Rightarrow n=50$