Question

Solve in positive integers $2xy-4x^2+12x-5y=11$

• A. $(4,11)$
• B. $(3,11)$
• C. $(3,10)$
• D. $(4,9)$

Answer 1

Answer: (B), (D)

The correct options are
B $(3,11)$
D $(4,9)$

Given $2xy-4x^2+12x-5y=11$

$4x^2-x(12+2y)+5y-11=0$

$x=\frac{(12+2y)\pm \sqrt{{(12+2y)}^2-16(5y+11)}}{2\times 4}$

$x=\frac{(12+2y)\pm \sqrt{144+4y^2+48y-80y-176}}{8}$

$x=\frac{(12+2y)\pm \sqrt{4y^2-32y-32}}{8}$

For positive integers

$4y^2-32y-32$ must be perfect square

$4(y^2-8y-8)$ must be perfect square

$(y^2-8y-8)$ must be perfect square

${(y-4)}^2-24=k^2\left(assume\right)$

${(y-4)}^2-k^2=24$

$(y-4-k)(y-4+k)=2\times 12or4\times 6$

Case 1

$(y-4-k)=2(y-4+k)=12$

Adding $2(y-4)=14$

$y=11$

Case 2

$(y-4-k)=4(y-4+k)=6$

Adding $2(y-4)=10$

$y=9$

For $y=11$

$x=\frac{12+2\times 11\pm \sqrt{100}}{8}=\frac{44}{8}or3$

For $y=9$

$x=\frac{12+2\times 9\pm \sqrt{4}}{8}=\frac{32}{8}=4or\frac{28}{3}$

As they are integers hence $(3,11)$ or$(4,9)$