Tìm các cặp số nguyên dương a,b sao cho: a^3-b^3+3(a^2-b^2)+3(a-b)=(a+1)(b+1)+25

Giải thích các bước giải:

$a^3-b^3+3(a^2-b^2)+3(a-b)=(a+1)(b+1)+25$ 

$\to (a^3+3a^2+3a+1)-(b^3+3b^2+3b+1)=(a+1)(b+1)+25$

$\to (a+1)^3-(b+1)^3=(a+1)(b+1)+25$

$\to (a+1)^3-3(a+1)(b+1)(a+1-(b+1))-(b+1)^3+3(a+1)(b+1)(a+1-(b+1))=(a+1)(b+1)+25$

$\to (a+1-b-1)^3+3(a+1)(b+1)(a-b)=(a+1)(b+1)+25$

$\to (a-b)^3+3(a+1)(b+1)(a-b)=(a+1)(b+1)+25$

Đặt $(a+1)(b+1)=y, (a-b)=x$

$\to x^3+3xy=y+25$

$\to y(3x-1)=25-x^3$

$\to y=\dfrac{25-x^3}{3x-1}$

$\to \dfrac{25-x^3}{3x-1}\in Z$

$\to \dfrac{675-(3x)^3}{3x-1}\in Z$

$\to \dfrac{674+1-(3x)^3}{3x-1}\in Z$

$\to \dfrac{674}{3x-1}\in Z$

$\to 3x-1\in U(674)=\{2,337,674,-674,-337,-2\}\to x\in\{1,225,-112,\}\to y\to a,b$