chứng minh rằng nếu $n$ là số nguyên dương thì : $5^n(5^n + 3^n) - 2^n(9^n + 11^n)$ chia hết cho $21$

Đáp án:

$[5^n(5^n+3^n)-2^n(9^n+11^n)]\vdots 21$

Giải thích các bước giải:

Đặt: $A=5^n(5^n +3^n)-2^n (9^n +11^n)$

$= 25^n + 15^n - 18^n -22^n\\25\equiv 1\pmod{3}\to 25^n\equiv 1\pmod{3}\\15^n\equiv 0\pmod{3}\\18^n\equiv 0\pmod{3}\\22\equiv 1\pmod{3}\to 22^n\equiv 1\pmod{3}\\\to A\equiv 1+0-0-1\equiv 0\pmod{3}\\\to A\vdots 3(1)$

Mặt khác:

$25\equiv 4\pmod{7}\to 25^n\equiv 4^n\pmod{7}\\15\equiv 1\pmod{7}\to 15^n\equiv 1\pmod{7}\\18\equiv 4\pmod{7}\to 18^n\equiv 4^n\pmod{7}\\22\equiv 1\pmod{7}\to 22^n\equiv 1\pmod{7}\\\to A\equiv  4^n -4^n+1-1\equiv 0\pmod{7}\\\to A\vdots 7(2)$

Ta thấy $(3;7)=1$ nên từ $(1),(2)\to A\vdots (3.7)\to A\vdots 21$