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I am in high school right now and I would like to learn how to approach this sort of problems. I think this is called a diophantine eqution. Thanks a bunch

This is what I deduces, just so you know the work done (though I doubt its useful)

$x|x(x-1)+y(y-1) \rightarrow x|y(y-1)$

$y|x(x-1)+y(y-1) \rightarrow y|x(x-1)$

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3 Answers 3

up vote 7 down vote accepted

HINT:

On rearrangement, $$x^2-x(1+y)+y^2-y=0$$ which is a Quadratic equation in $x$

The discriminant is $(1+y)^2-4(y^2-y)=1-3y^2+6y=4-3(y-1)^2$

For real $x, 4-3(y-1)^2\ge0\implies (y-1)^2\le\frac43\implies -\frac2{\sqrt3}\le y-1\le \frac2{\sqrt3}$

As $\frac2{\sqrt3}<2, -2<y-1<2\iff-1<y<3$

As $y$ is integer, $0\le y\le2$

Test for each value

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First note that $2|xy|\le x^2+y^2$, then by the equation we must have $|x|+|y|\ge \frac{1}{2}(x^2+y^2)$, which imply that $(x\pm 1)^2+(y\pm1)^2\le2$. The only thing left is to check which $(x,y)$ satisfying this condition.

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why $|x|+|y| \geq \frac{1}{2}(x^2+y^2)$ –  Jorge Fernández Aug 3 '13 at 16:19
    
@Omnitic $|x|+|y|\ge x+y=x^2+y^2-xy\ge\frac{1}{2}(x^2+y^2)$ –  asatzhh Aug 3 '13 at 16:26

This particular curve is an ellipse, which you can see by plugging into wolfram alpha or by expressing it in standard form, which is $x^2-xy+y^2-x-y=0$ and considering the discriminant, which in this case is $-3$.

This particular ellipse has integer solutions $(0,0)$, $(1,0)$, $(0,1)$, $(1,2)$, $(2,1)$, $(2,2)$ which can be found by inspection. There are no more because the ellipse is only so big.

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How does inspection work? –  Jorge Fernández Aug 3 '13 at 16:10
    
Inspection works by trying all values of $x,y$ between $-1$ and $3$, say. If you look at the plot on wolfram alpha you'll see the ellipse isn't so big. –  vadim123 Aug 3 '13 at 16:11
1  
Is there an algebraic way to approach this other than using computing power to plot the equation? –  Jorge Fernández Aug 3 '13 at 16:12
    
There are ways to plot conic sections by hand, which you ought to be learning in high school. The methods are a bit tedious if there's an $xy$ term, such as here. –  vadim123 Aug 3 '13 at 16:13
1  
you missed (2,2) –  Jorge Fernández Aug 6 '13 at 19:03

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