Question

Solve in positive integers:
$y^2-4xy+5x^2-10x=4$.

Answer 1

Since we have 4xy term the equation can be written as $(y-2x{)}^2+...$

$=>y^2-4xy+4x^2+x^2-10x=4$

adding 25 on both sides

$=>(y-2x{)}^2+(x-5{)}^2=29$

$29=25+4=5^2+2^2$

by comparision if $a^2=b^2$ then $a=+b$ or $a=-b$

case 1: $x-5=2;y-2x=5$;

$=>x=7;y=19$

case 2: $x-5=-2;y-2x=5$;

$=>x=3;y=11$

case 3: $x-5=2;y-2x=-5$;

$=>x=7;y=9$

case 4: $x-5=-2;y-2x=-5$;

$=>x=3;y=1$

$(x,y)=(7,19);(3,11);(7,9);(3,1)$

case 5: $x-5=5;y-2x=2$;

$=>x=10;y=22$

case 6: $x-5=-5;y-2x=2$;

$=>x=0;y=2$ but given $x,y$ are positive.Hence not a solution

case 7: $x-5=5;y-2x=-2$;

$=>x=10;y=18$

case 8: $x-5=-5;y-2x=-2$;

$=>x=0;y=-2$ but given $x,y$ are positive.Hence not a solution

$(x,y)=(7,19);(3,11);(7,9);(3,1);(10,18);(10,22)$