by Da-iCE
Last Updated July 12, 2019 06:20 AM

Find all the integer solutions involving the following Diophantine-equation
$$\frac{x+ y}{x^{2}+ y^{2}+ xy}= \frac{5}{19}$$
By **W**|**A** I found that $2$ and $3$ are the solutions of the O-p, my observation is
$$5\left ( 5(x^{2}+ y^{2}+ xy)- 19(x+ y) \right )+ (3x- 2y)(3y- 2x)= 19(x+ y- 5)(x+ y)$$
$$6\left ( 5(x^{2}+ y^{2}+ xy)- 19(x+ y) \right )+ 5(3x- 2y)(3y- 2x)= 19(5xy- 6x- 6y)$$
Then, how can I combine them with $5\mid x+ y$ and $19\mid x^{2}+ y^{2}+ xy$, thanks

Hint: Simplifying your given equation we obtain $$-5x^2-5xy-5y^2+19x+19y=0$$ solving for $y$ we get $$y_{1,2}=-\frac{x}{2}+\frac{19}{10}\pm\frac{\sqrt{-75x^2+190x+361}}{10}$$ For the discriminant we have $$-75x^2+190x+361\geq 0$$ This is $$-\frac{19}{15}\le x\le \frac{19}{2}$$ Now you can find the integer solutions for $x$.

$$5x^2+x(5y-19)+5y^2-19y=0$$

$\dfrac{x+y}5=\dfrac{x^2+xy+y^2}{19}=c$(say)

$(x+y)^2-xy=19c\implies xy=25c^2-19c$

$x,y$ are the roots of $$t^2-5ct+25c^2-19c=0$$

The discriminant $$(5c)^2-4(25c^2-19c)=c(76-75c)$$

$0<c\le\dfrac{76}{75}$

If $c$ is an integer, $c=1$

Updated August 11, 2017 08:20 AM

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