$$2x^2+5x+y^2=19$$
Don't know how to approach the problem. Similar equations required factoring after the completing a square or a similar trick. I don't see the possibility of that here though. Hints? Answers?
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I don't see how this helps.. – John Doe May 27 '15 at 22:49
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Actually, multiply through by $8,$ complete the square in $x,$ it is an ellipse that does have some integer points. http://www.printablepaper.net/category/graph – Will Jagy May 27 '15 at 22:50
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There are only a relatively small number of possibilities to work through. $(x,y) = {(2,\pm1),(-3,\pm4)}$ – Joffan May 27 '15 at 23:13
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Isn't this just an ellipse? That gives you a finite number of possibilities. – DanielV May 28 '15 at 03:48
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Note that $$\begin{align}19-2x^2-5x=y^2\ge 0&\Rightarrow 2x^2+5x-19\le 0\\&\Rightarrow \frac{-5-\sqrt{177}}{4}\le x\le\frac{-5+\sqrt{177}}{4}\\&\Rightarrow -4\le x\le 2\end{align}$$
mathlove
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Are these types of exercises often solved using this method or is it just handy to use in this situation? – John Doe May 27 '15 at 22:53
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1Because of the geometry (ellipse) there are only finitely many candidates. So it is reasonable to put bounds on the candidates, and then test them all. – André Nicolas May 27 '15 at 22:56
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@JohnDoe: As André Nicolas comments, you can use the method for example when you have $ax^2+bx+cy^2=d$ where $a,b$ are positive. – mathlove May 27 '15 at 23:00
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The equation with small numbers allows easy calculations. As a simple quadratic equation we have $4x = -5 ±\sqrt{177-8y^2}$. There are just four possibilities: y = ±1, ±2, ±3, ±4 and only ±1 and ±4 are good. Thus $4x = -5 ±13$ and $4x = -5 ±7$. Finally the only solutions are (x, y) = (2, ±1) and (-3, ±4)
Piquito
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