Determine the pairs of natural numbers $(x,y)$ for which the relation $2x^2+5y^2-11xy=-25$ occurs

Question

the question

Determine the pairs of natural numbers $(x,y)$ for which the relation $2x^2+5y^2-11xy=-25$ occurs.

my idea

I tried grouping them in a whole perfect square or I tried using formulas such as

$$a^2+2ab+b^2=(a+b)^2$$

$$a^2-2ab+b^2=(a-b)^2$$

As I didn't got to a convenient form I decided to write it as:

$2x^2+5y^2-11xy+25=0$ and thought of using the quadric formula.

$x= \frac{11y +- \sqrt{81y^2-200}}{4}$

I don't know what to do forward. Hope one of you can help me! Thank you!

Answer 1

As said, factor $$2x^2+5y^2-11xy=(x-5y)(2x-y)$$ We have the following cases:

1. $-25=-1\cdot 25$. Solve the systems: $$\begin{cases}x-5y=-1\\2x-y=25\end{cases}\:\Longrightarrow\:\boxed{(x,y)=(14,3)}$$ and $$\begin{cases}x-5y=25\\2x-y=-1\end{cases}\:\Longrightarrow\:(x,y)=\left(-\frac{10}{3},-\frac{17}{3}\right)$$ We can't accept this last solution, since $x$ and $y$ are not natural numbers.

2. $-25=-5\cdot 5$. Solve: $$\begin{cases}x-5y=-5\\2x-y=5\end{cases}\:\Longrightarrow\:(x,y)=\left(\frac{10}{3},\frac{5}{3}\right)$$ and $$\begin{cases}x-5y=5\\2x-y=-5\end{cases}\:\Longrightarrow\:(x,y)=\left(-\frac{10}{3},-\frac{5}{3}\right)$$ None of the solutions can be accepted.

So, the only solution is: $$\boxed{(x,y)=(14,3)}$$

Further details: At first glance it seems we have to analyze also the case $-25=-25\cdot 1$, so that we have to solve the systems: $$\begin{cases}x-5y=-25\\2x-y=1\end{cases}$$ and $$\begin{cases}x-5y=1\\2x-y=-25\end{cases}$$ Observing that they only have opposite signs compared to the first case we analyzed, the solutions are, respectively, $\left(\frac{10}{3},\frac{17}{3}\right)$ and $(-14,-3)$. Neither of them can be accepted, since the solutions are not natural numbers.

Edit 2: How did I obtain the factorization $2x^2+5y^2-11xy=(x-5y)(2x-y)$? I just looked for a decomposition of the form $$2x^2+5y^2-11xy=(x+ay)(2x+by)$$ and solved for some $a$ and $b$...

Answer 2

Hint: when problem-solving equations of this kind where solutions are limited to the natural numbers, often you want to look for convenient factorisations so you can apply number theory ideas.

In this case, you might like to consider if $2x^2+5y^2-11xy$ can be factorised.

Answer 3

Hint: continue with what you found:

$\Delta=81y^2-200\geq 0$

For $y=3$ we get $\Delta=23^2$

$x=\frac{11\times 3\pm 23}4$

The positive solution is $x=14$. You can find more $y$ for a perfect square of $\Delta $ an for x.In fact we may write:

$81y^2-200=k^2$

This is a Pell like equation which has standard method of solution.

You can also factorize LHS as:

$(2x+y)(5y+X)=-25$

But I could not find integral solutions by this way.