1
$\begingroup$

Find all ordered pairs (x,y) of positive integers x, y such that $x^2+4y^2=(2xy−7)^2$. I get the ordered pair (3,2) as the only solution and I was wondering if there could be anything else.

If someone has the solution for this I would greatly appreciate.

$\endgroup$
0
$\begingroup$

I have a solution.

Solution:

$$x^2 +4y^2=(2xy-7)^2$$ $$(x+2y)^2=4(xy)^2-24xy+49$$ Manipulating it into a more useful form: $$(x+2y)^2=4(xy)^2-24xy+49=4(xy)(xy-6)+49$$ Think $(xy)$ to be equal to $z$ and observe that in the given we just need to find positive integers hence possible solution of $xy$ could be $\pm6$ or $0$.As $0$ Doesn't satisfy the given equation and only positive integers are to be considered $6$ is the product of $xy$ and little observation shows only $(3,2)$ is the only pair possible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.