# Diophantine equation - II

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.

## 1 Answer

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.