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 \(\left(5;8;10\right)=1\)

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

a) Với \(n\in N\Rightarrow2^{4n}-1=16^n-1=\left(16-1\right).\left(16^{n-1}+16^{n-2}+...+1\right)\)

\(=15.\left(16^{n-1}+16^{n-2}+...+1\right)⋮15\)

b) Với \(n\in N\Rightarrow16^n-15n-1=\left(16^n-1\right)-15n\)

mà \(\left(16^n-1\right)⋮15\left(cma\right);15n⋮15\)

\(\Rightarrow16^n-15n-1⋮15\)