Given positive integers, if $$ x^2+y^2+4y-96=0 $$ , find the value of $$ xy $$ .

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Uniteasy Admin · Accepted Answer

$$ \because x^2+y^2+4y-96=0 $$
$$ 	herefore  x^2=100-(y+2)^2>0 $$
 If $$ y=1 $$ , $$ 100-(1+2)^2=91 $$ , not a square number.
 If $$ y=2 $$ , $$ 100-(2+2)^2=84 $$ , not a square number.
 If $$ y=3 $$ , $$ 100-(3+2)^2=75 $$ , not a square number.
 If $$ y=4 $$ , $$ 100-(4+2)^2=64 $$ , $$ x=8 $$
 If $$ y=5 $$ , $$ 100-(5+2)^2=51 $$ , not a square number.
 If $$ y=6 $$ , $$ 100-(6+2)^2=36 $$ , $$ x=6 $$
 If $$ y=7 $$ , $$ 100-(7+2)^2=19 $$ , not a square number.
 If $$ y=8 $$ , $$ 100-(8+2)^2=0 $$ , $$ x=0 $$
$$ 	herefore y=4 $$ or $$ 6 $$ , $$ xy=32 $$ or $$ 36 $$

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