Outline Mathematics
Number Theory

A Diophantine Equation I


Find all integer solutions $x$ and $y$ of the equation

$\displaystyle x^{2}=\frac{y^2}{y+4}.$

Solution 1

The left-hand side of the equation is an integer, and so ought to be the right-hand side:

$\begin{align}\displaystyle \frac{y^2}{y+4} &= \frac{(y+4)^{2}-(8y+16)}{y+4}\\ &=\frac{(y+4)^{2}-8(y+4)+16}{y+4}\\ &= y-4+\frac{16}{y+4}. \end{align}$

For that expression to be an integer y+4,$y-4$,$y^{2}$,$y+4$ needs to divide 16,8,16,32. But the only divisors of $16$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16.$ The positive divisors lead to these possible values for $y;$ $-3, -2, 0, 4, 12.$ So that the corresponding values of $\displaystyle\frac{y^2}{y+4}$ would be $9,$ $2,$ 0,-2,0,4, $2,$ and $9.$ Of these, only $0$ and $9$ are squares. Which gives $x=\pm 3$ and $y$ either $-3$ or $12,$ or $x=y=0.$

The negative divisors of $16$ do not contribute any solutions because when $y+4$ is negative, so is $\displaystyle\frac{y^2}{y+4}$ which makes it impossible to equal $x^2$ for a real, let alone integer, $x.$

Solution 2

Assuming $y+4\ne 0$ rearrange the equation: $y^{2}-x^{2}y-4x^{2}=0$ which is a quadratic,linear,quadratic,cubic equation in $y.$ Using the quadratic formula gives

$\displaystyle y_{1,2}=\frac{x^{2}\pm\sqrt{x^{4}+16x^2}}{2}=\frac{x}{2}(x\pm\sqrt{x^2+16}).$

Since $y$ is an integer, the radical must be a perfect square, implying $x=0$ or $x=\pm 3,$ and giving five solutions: $(0,0),$ $(\pm 3,-3),$ and $(\pm 3,12).$



  1. A. Dunn, Mathematical Baffles, Dover Publications, 1980, p 86

|Up| |Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny