Ta có: A = 20ⁿ + 16ⁿ - 3ⁿ - 1

Do n chẵn => n = 2k

Khi đó: A = 20^2k + 16^2k - 3^2k - 1

A = (20^2k - 1) + (256^k - 9^k) 

Do 20^2k - 1 \(⋮\)(20 - 1) = 19

 256^k - 9^k \(⋮\)(256 - 9) = 247 \(⋮\)19

=> A \(⋮\)19 (1)

Mặt khác, ta lại có: 

A = 20^2k + 16^2k - 3^2k - 1 = (20^2k - 3^2k) + (256^k - 1)

Do 20^2k - 3^2k \(⋮\)(20 - 3) = 17

256^k - 1 \(⋮\)(256 - 1)= 255 \(⋮\)17

=> A  \(⋮\)17 (2)

Mà (17; 19) = 1 => A \(⋮\)17.19 = 323 (đpcm)

Vì n chẵn 

Đặt n = 2k (k \(\inℕ\))

Khi đó A = 20ⁿ + 16ⁿ - 3ⁿ - 1

= 20^2k + 16^2k - 3^2k - 1 

= 400^k + 256^k - 9^k - 1

= (400^k - 1) + (256^k - 9^k)

= (400 - 1)(400^{k - 1} + 400^{k - 2} + ... + 1) + (256 - 9)(256^{k - 1} + 256^{k - 2}.9 + ... + 9^{k - 1})

= 399(400^{k - 1} + 400^{k - 2} + ... + 1) + 247(256^{k - 1} + 256^{k - 2}.9 + ... + 9^{k - 1})

= 19.21.(400^{k - 1} + 400^{k - 2} + ... + 1) + 19.13(256^{k - 1} + 256^{k - 2}.9 + ... + 9^{k - 1})

= 19.(21.(400^{k - 1} + 400^{k - 2} + ... + 1) + 13(256^{k - 1} + 256^{k - 2}.9 + ... + 9^{k - 1})) \(⋮\)19 (1)

Lại có A = 400^k + 256^k - 9^k - 1 

= (400^k - 9^k) + (256^k - 1)

= (400 - 9)(400^{k - 1} + 400^{k - 2}.9 + .... + 9^{k - 1}) + (256 - 1)(256^{k - 1} + 256^{k - 2} + .... + 1)

= 391(400^{k - 1} + 400^{k - 2}.9 + .... + 9^{k - 1}) + 255(256^{k - 1} + 256^{k - 2} + .... + 1)

= 17.23(400^{k - 1} + 400^{k - 2}.9 + .... + 9^{k - 1}) + 17.15(256^{k - 1} + 256^{k - 2} + .... + 1)

= 17.(23(400^{k - 1} + 400^{k - 2}.9 + .... + 9^{k - 1}) + 15(256^{k - 1} + 256^{k - 2} + .... + 1)) \(⋮\)17 (2)

Lại có ƯCLN(17;19) = 1 (3)

Từ (1)(2)(3) => A \(⋮17.19=323\)(ĐPCM)

Đặt: n⁴ + 2n³ + 2n²+ n + 7 = k² (k \(\in\)N)

<=> (n² + n)² + (n² + n) + 7 = k²

<=> 4(n² + n)² + 4(n² + n) + 28 = 4k²

<=> 4k² - (2n² + 2n + 1)² = 27

<=> (2k - 2n² - 2n - 1)(2k + 2n² + 2n + 1) = 27

Do 2k + 2n² + 2n + 1 > 2k - 2n² - 2n - 1

Lập bảng

 (tự tính)

Đặt \(B=4^{1975}+4^{1974}+...+4^2\)

\(\Rightarrow4B=4^{1976}+4^{1975}+...+4^3\)

\(\Rightarrow4B-B=\left(4^{1976}+4^{1975}+...+4^3\right)-\left(4^{1975}+4^{1974}+...+4^2\right)\)

hay \(3B=4^{1976}-4^2\)

\(\Rightarrow B=\frac{4^{1976}-4^2}{3}\)

\(\Rightarrow A=75\left(B+5\right)+25\)

\(=75\left(\frac{4^{1976}-4^2}{3}+5\right)+25\)

\(=25.\left(4^{1976}-16\right)+375+25\)

\(=25.4^{1976}-400+400\)

\(=25.4^{1976}⋮4^{1976}\left(đpcm\right)\)

Ta có : \(A=7+7^2+7^3+...+7^{4k}\)

\(=\left(7+7^2+7^3+7^4\right)+...+\left(7^{4k-3}+7^{4k-2}+7^{4k-1}+7^{4k}\right)\)

\(=\left(7+7^2+7^3+7^4\right)+...+7^{4k-4}\left(7+7^2+7^3+7^4\right)\)

\(=\left(7+7^2+7^3+7^4\right)\left(1+...+7^{4k-4}\right)\)

\(=2800\left(1+...+7^{4k-4}\right)\)

\(=350.8\left(1+...+7^{4k-4}\right)⋮8\)

\(\Rightarrow A⋮8\left(1\right)\)

Ta lại có : \(A=7+7^2+7^3+...+7^{4k}\)

\(\Rightarrow7A=7^2+7^3+7^4+...+7^{4k+1}\)

\(\Rightarrow7A-A=\left(7^2+7^3+7^4+...+7^{4k+1}\right)-\left(7+7^2+7^3+....+7^{4k}\right)\)

hay \(6A=7^{4k+1}-7=7\left(7^{4k}-1\right)\)

Vì \(7\equiv2\left(mod5\right)\)\(\Rightarrow7^{4k}\equiv2^{4k}=16^k\left(mod5\right)\)

mà \(16\equiv1\left(mod5\right)\)\(\Rightarrow16^k\equiv1^k=1\left(mod5\right)\)

\(\Rightarrow7^{4k}\equiv1\left(mod5\right)\)

\(\Rightarrow7^{4k}-1⋮5\left(\cdot\right)\)

\(\Rightarrow7\left(7^{4k}-1\right)⋮5\)

\(\Rightarrow6A⋮5\)

Nhưng \(\left(6;5\right)=1\)

\(\Rightarrow A⋮5\left(2\right)\)

Ta lại có tiếp : \(7\equiv1\left(mod2\right)\)

\(\Rightarrow7^{4k}\equiv1^{4k}=1\left(mod2\right)\)

\(\Rightarrow7^{4k}-1⋮2\left(\cdot\cdot\right)\)

Từ \(\left(\cdot\right)\), \(\left(\cdot\cdot\right)\) và \(\left(2;5\right)=1\): \(\Rightarrow7^{4k}-1⋮10\)

\(\Rightarrow7\left(7^{4k}-1\right)⋮10\)

\(\Rightarrow6A⋮10\)

Nhưng \(\left(6;10\right)=1\)

\(\Rightarrow A⋮10\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\)và \(\left(5;8;10\right)=1\)

\(\Rightarrow A⋮400\left(đpcm\right)\)