How many integer pairs $$(x,y)$$ satisfy $$x^{2}$$$$+$$$$4y^{2}$$$$-$$$$2xy$$$$-$$$$2x$$$$-$$$$4y$$$$-$$$$8$$ $$=$$ $$0$$? Also,how can we solve this using the method of 'completing the squares'?

Note by Bhargav Das
4 years, 9 months ago

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Completing the square in $$x$$, and then completing the square in $$y$$, this equation becomes $\begin{array}{rcl} (x-y-1)^2 + 3y^2 - 6y - 9 & = & 0 \\ (x-y-1)^2 + 3(y-1)^2 & = & 12 \end{array}$ Thus $$|y-1| = 1$$ or $$2$$, and we obtain the six solutions $$(4,3)$$, $$(0,-1)$$, $$(6,2)$$, $$(4,0)$$, $$(0,2)$$ and $$(-2,0)$$.

- 4 years, 9 months ago

Thanks a lot.

- 4 years, 9 months ago