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 May 3rd, 2014, 04:38 AM #1 Newbie   Joined: May 2014 From: UK Posts: 4 Thanks: 0 Parabolas Hello there, I've got the following problem that I require some help with: x^2-3y-4=0, I need to find all values of (x,y) that are integers. Has anybody got any ideas? Ty in advance, Bearz
 May 3rd, 2014, 05:54 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra $$y = \frac{1}{3}(x - 2)(x + 2)$$ Thus, given $x \in \mathbb{Z}$, $x - 2 = 3k$ and $x+2 =3k$ ($k \in \mathbb{Z}$) give $y \in \mathbb{Z}$. So, all $x = 3k \pm 2$ are integer solutions which is the same as saying all integers not divisible by three. We can also investigate the $y$ values. $$y = k(3k \pm 4) = 3k^2 \pm 4k$$ Thanks from Olinguito and Bearz Last edited by v8archie; May 3rd, 2014 at 06:05 AM.

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