Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$

Question

For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$ Find the biggest and smallest value for $x-y$.

How do you proceed with such a question? Are their formulas or something for that type of equation?

I'd appreciate any help.

Answer 1

Notice that $y^2-1$ is adivisor of $240$; also we know that $-1 \leq y^2-1$, by checking throuh all divisors of $240$, we get the folowing possibilities for $y^2-1$:

• $y^2-1=-1$ but there is no pair in this case.

• $y^2-1=0$ but there is no pair in this case.

• $y^2-1=3$ and $x^2+x=80$; but notice that there is no solution to the equation $x^2+x=80$; so there is no pair $(x,y)$ for the main equation, in this case.

• $y^2-1=8$ and $x^2+x=30$; which gives $(x,y)=(5,\pm3)$ and $(x,y)=(-6,\pm3)$.

• $y^2-1=15$ and $x^2+x=16$; but notice that there is no solution to the equation $x^2+x=16$; so there is no pair $(x,y)$ for the main equation, in this case.

• $y^2-1=24$ and $x^2+x=10$; but notice that there is no solution to the equation $x^2+x=10$; so there is no pair $(x,y)$ for the main equation, in this case.

• $y^2-1=48$; and $x^2+x=5$; but notice that there is no solution to the equation $x^2+x=5$; so there is no pair $(x,y)$ for the main equation, in this case.

• $y^2-1=80$; and $x^2+x=3$; but notice that there is no solution to the equation $x^2+x=3$; so there is no pair $(x,y)$ for the main equation, in this case.

• $y^2-1=120$ and $x^2+x=2$; which gives $(x,y)=(1,\pm11)$ and $(x,y)=(-2,\pm11)$.

Answer 2

Here is a table of whole numbers $x$ so that $x^2+x\mid240$: $$ \begin{array}{c|c} x&1&2&3&4&5&15\\\hline x^2+x&\color{#090}{2}&6&12&20&\color{#090}{30}&240 \end{array} $$ For $x$ to work, we need $\frac{240}{x^2+x}+1=y^2$:

$\frac{240}{2}+1=11^2$
$\frac{240}{6}+1=41$
$\frac{240}{12}+1=21$
$\frac{240}{20}+1=13$
$\frac{240}{30}+1=3^2$
$\frac{240}{240}+1=2$

The choices we have are $\{x=1,y=11\}$ and $\{x=5,y=3\}$. Thus, the smallest is $x-y=-10$ and the largest is $x-y=2$.

Answer 3

$(x^2+x)(y^2-1)=240$ implies that $y = \sqrt{1 + \dfrac{240}{x(x+1)}}$

Clearly both $x$ and $x+1$ need to be divisors of $240$.

These are the divisors of $240$

\begin{align} 1 &\quad 240 \\ 2 &\quad 120 \\ 3 &\quad 80 \\ 4 &\quad 60 \\ 5 &\quad 48 \\ 6 &\quad 40 \\ 8 &\quad 30 \\ 10 &\quad 24 \\ 12 &\quad 20 \\ 15 &\quad 16 \end{align}

For $x = 1,2,3,4,5,15$ we get $y = 11, \sqrt{41}, \sqrt{21}, \sqrt{13}, 3, \sqrt 2$

So the minimum value of $x-y$ is $1-11 = -10$ and the maximum value of $x-y$ is $5-3 = 2$