Please help in this N.T Prob.

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'?

#NumberTheory #MathProblem #Math

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $(-2,0)$ .

Mark Hennings - 7 years, 8 months ago

Thanks a lot.

Bhargav Das - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$