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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!

3 Answers3

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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.

cambridgecircus
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  • Is it ok to write it as $x^2+4y^2-4xy+x^2+y^2-2xy-5xy=-25= (x-2y)^2+(x-y)^2-5xy$? –  Feb 08 '24 at 08:19
  • You haven't understood the hint. – cambridgecircus Feb 08 '24 at 08:24
  • Can you please be more explicit? –  Feb 08 '24 at 08:41
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    @RyanPamMunoz Just observe that $2x^2+5y^2-11xy=(x-5y)(2x-y)$... Then what are the factors of $-25$? – user773458 Feb 08 '24 at 08:45
  • @user773458 Are they $1,5,25,-1,-5,-25$? –  Feb 08 '24 at 08:47
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    @user773458: in order to get to that factorisation, did you just tried $y=1$, solved the second-degree equation in $x$ ($x=5$ and $x=\frac{1}{2}$ and used this to write the equation as $(x-5) \cdot (x-\frac{1}{2})$, multiplied by 2 and replace $1$ in the solution back to $y$? (This is very naïve, I know, but it might help to understand the way of working. – Dominique Feb 08 '24 at 09:03
  • @Dominique I looked for a factorization of the form $2x^2+5y^2-11xy=(x+ay)(2x+by)$ and solved form some $a$ and $b$... – user773458 Feb 08 '24 at 09:36
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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.

sirous
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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 $b$...

user773458
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