Obtain all ordered pair of integers $(x,y)$ such that $x(x + 1) = y(y + 1)(y + 2)(y + 3)$
I'm getting 8,
(0, 0), (0, -1), (0, -2), (0, -3) (-1, 0), (-1, -1), (-1, -2), (-1, -3)
Please confirm my answer.
Obtain all ordered pair of integers $(x,y)$ such that $x(x + 1) = y(y + 1)(y + 2)(y + 3)$
I'm getting 8,
(0, 0), (0, -1), (0, -2), (0, -3) (-1, 0), (-1, -1), (-1, -2), (-1, -3)
Please confirm my answer.
Hint: It is easily proved that the product of four consecutive integers, plus $1$, is a perfect square. But $x(x+1)+1$ is hardly ever a perfect square!
Added: To prove that $y(y+1)(y+2)(y+3)+1$ is a perfect square, note that $y(y+1)(y+2)(y+3)=y(y+3)(y+1)(y+2)=(y^2+3y)(y^2+3y+2)=z(z+2),$ where $z=y^2+3y$. And clearly $z(z+2)+1=(z+1)^2$.
HINT Assume $x>0$. Then this question along with the fact that $x^2 < x^2 + x + 1 < (x+1)^2$ should do the job. Argue similarly for $x \leq 0$