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Find all integers $x$ such that $x^2+3x+24$ is a perfect square.

My attempt:

$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$

Now, do I find solution treating cases? But that doesn't seem very easy. Please help.

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Complete the square to get $(x+3/2)^2 + 87/4$. We want this to be a square itself, so $$(x+3/2)^2 + 87/4 = k^2.$$

This is the same as $$(2x+3)^2 + 87 = 4k^2.$$

Now consider the difference of squares $(2k - 2x - 3)(2k + 2x + 3) = 87 = 3\cdot 29$. Now there are only a few cases to check.

  1. $2k + 2x + 3 = 3$ and $2k - 2x - 3 = 29$ means $4k = 32$, $k = 8$.
  2. $2k + 2x + 3 = 87$ and $2k - 2x - 3 = 1$ means $4k = 88$, $k = 22$.

Do these give you any integer solutions for $x$?

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    $\begingroup$ If $a\cdot b=87$ then you also have $a=\pm 87,b=\mp 1$ (and the inverse). In the cases you have considered you have to take all the possible signs. $\endgroup$ – 111 Feb 1 '15 at 13:17
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Best way is....

$$x^2 + 3x + 24 = k^2 \Rightarrow x^2 + 3x + 24-k^2 = 0\\$$

$$\triangle = b^2 - 4ac = 3^2 - 4(1)(24-k^2) = 9 - 96 + 4k^2 = 4k^2 - 87 = n^2\\$$

$$\to 4k^2 - n^2 = (2k-n)(2k+n) = 87 \\$$

$$\to 2k+n = 29, 2k-n = 3 \to 4k = 32 \to k = 8$$. Can you finish it?

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  • $\begingroup$ wait, I do not understand why $4k^2-87=n^2$? $\endgroup$ – Swadhin Feb 1 '15 at 7:18
  • $\begingroup$ it has to be a perfect square for the equation to have integer roots. $\endgroup$ – DeepSea Feb 1 '15 at 7:31
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Hint Consider $$4k^2=(2x+3)^2+87$$

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  • $\begingroup$ I have doubts about it, would you please elaborate? $\endgroup$ – Swadhin Feb 1 '15 at 6:57
  • $\begingroup$ @Swadhin Ask yourself this: When can two perfect squares differ by $87$? $\endgroup$ – Slade Feb 1 '15 at 7:40
  • $\begingroup$ When the squares are $44$ and $43$ $\endgroup$ – Swadhin Feb 1 '15 at 8:00
  • $\begingroup$ @Swadhim: That's not the only possibility if $(x+y)(x-y) = 87$. $\endgroup$ – TonyK Feb 1 '15 at 14:04

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