Tìm nghiệm nguyên của phương trình x^2+x+2009=y^2

Đáp án: $(y,x)\in\{(2009, 2008), (403,400), (-2009, 2008), (-4003,-4001)\}$

Giải thích các bước giải:

Ta có:
$x^2+x+2009=y^2$
$\to 4x^2+4x+8036=4y^2$
$\to (4x^2+4x+1)+8035=(2y)^2$
$\to (2x+1)^2+8035=(2y)^2$
$\to (2y)^2-(2x+1)^2=8035$
$\to (2y-2x-1)(2y+2x+1)=8035$
$\to (2y-2x-1, 2y+2x+1)$ là cặp nghiệm của $8035$
Mà $2y-2x-1 + 2y+2x+1=4y\quad\vdots\quad 4$
$\to (2y-2x-1, 2y+2x+1)\in\{(1,8035), (8035,1), (5,1607), (1607,5),(-1,-8035), (-8035,-1), (-5,-1607), (-1607,-5)\}$

$\to (2y-2x, 2y+2x)\in\{(2,8034), (8036,0), (6,1606), (1608,4),(0,-8036), (-8034,-2), (-4,-1608), (-1606,-6)\}$

$\to (y-x, y+x)\in\{(1,4017), (4017,0), (3,803), (803,2),(0,-4017), (-4017,-1), (-2,-8004), (-8004,-3)\}$

Vì $y-x+y+x$ chẵn

$\to (y-x, y+x)\in\{(1,4017),  (3,803), (-4017,-1), (-2,-8004)\}$

$\to (2y,2x)\in\{(4018, 4016), (806,800), (-4018, 4016), (-8006,-8002)\}$

$\to (y,x)\in\{(2009, 2008), (403,400), (-2009, 2008), (-4003,-4001)\}$