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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.

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    Thank you sir Andre, sir Marvis and others.2012-06-27

2 Answers 2

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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 $z(z+2)+1=(z+1)^2$.

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    I can think of a couple of easy cases2012-06-27
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    @Henry: Thanks, I had made not a good choice of words.2012-06-27
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    x = 0, -1 are the only cases when x(x+1)+1 is a perfect square, right sir?2012-06-27
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    I'm getting 4 ordered pairs namely, (0,0), (0, -1), (0, -2), (0, -3). Please confirm my answer.2012-06-27
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    You mentioned $x=-1$ above, but then forgot about it.2012-06-27
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    Sorry sir @AndréNicolas I'm getting 8, (0, 0), (0, -1), (0, -2), (0, -3) (-1, 0), (-1, -1), (-1, -2), (-1, -3)2012-06-27
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    @Bazinga: No problem! You know what's going on, that's what's important.2012-06-27
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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 $x \leq 0$