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

Question

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.

score 5 · Answer 1 · edited Aug 20 '16 at 05:57

5

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?

edited Aug 20 '16 at 05:57

William Phoenix

answered Feb 01 '15 at 07:09

DeepSea

wait, I do not understand why $4k^2-87=n^2$? – Swadhin Feb 01 '15 at 07:18
it has to be a perfect square for the equation to have integer roots. – DeepSea Feb 01 '15 at 07:31

score 5 · Answer 2 · edited Feb 01 '15 at 13:00

5

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.

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

edited Feb 01 '15 at 13:00

N. F. Taussig

answered Feb 01 '15 at 07:11

Chris Brooks

1

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. – 111 Feb 01 '15 at 13:17

score 4 · Answer 3 · answered Feb 01 '15 at 06:53

4

Hint Consider $$4k^2=(2x+3)^2+87$$

answered Feb 01 '15 at 06:53

Tien Kha Pham

I have doubts about it, would you please elaborate? – Swadhin Feb 01 '15 at 06:57
@Swadhin Ask yourself this: When can two perfect squares differ by $87$? – Andrew Dudzik Feb 01 '15 at 07:40
When the squares are $44$ and $43$ – Swadhin Feb 01 '15 at 08:00
@Swadhim: That's not the only possibility if $(x+y)(x-y) = 87$. – TonyK Feb 01 '15 at 14:04

3 Answers3