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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.
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.
