the equation $9x^2 +2xy +y^2 -92x - 20y +244 =0$

Question

lf $x$ and $y$ are two real numbers connected by the equation $9x^2 +2xy +y^2 -92x - 20y +244 =0$ also if $m$ is the number of integral values of $x$ and $n$ be the number of integral values of $y$ which satisfy above relation then $mn =$?

Answer 1

$$9x^2+2xy+y^2-92x-20y+244=0$$ $$8x^2+x^2+2xy+y^2-72x-20x-20y+244=0$$ $$8x^2+(x+y)^2-72x-20(x+y)+244=0$$ $$8x^2-72x+144+(x+y)^2-20(x+y)+100=0$$ $$8(x^2-9x+18)+(x+y-10)^2=0$$ $$8(x-3)(x-6)+(x+y-10)^2=0$$ Then, note that $(x+y-10)^2\geq0$, which means that $(x-3)(x-6)\leq0$. This gives 4 possible values for x: 3,4,5,6. For each of these values, we can isolate the number of possible values for y.

When $x=3$, $8(x-3)(x-6)=0$. Therefore, $(x+y-10)^2=(y-7)^2=0$, which means that $y=7$ is the only possible value.

When $x=4$, $8(x-3)(x-6)=-16$. Therefore, $(x+y-10)^2=(y-6)^2=16$, which means that $y=10$ or $y=2$.

When $x=5$, $8(x-3)(x-6)=-16$. Therefore, $(x+y-10)^2=(y-5)^2=16$, which means that $y=9$ or $y=1$.

When $x=6$, $8(x-3)(x-6)=0$. Therefore, $(x+y-10)^2=(y-4)^2=0$, which means that $y=4$ is the only possible value.

In total, we have four possible integer values for x and six possible integer values for y. The answer to your question is therefore 24

Answer 2

Hint: We have $$ 9x^2 +2xy +y^2 -92x - 20y +244=9(x-5)^2+(2x + y - 19)(y - 1). $$