Finding integral roots for $ f(x,y)$ , the idea of Solution is credit to teacher Imad Zak

Exercise:

Let $f\left( x,y \right)={{x}^{2}}+{{y}^{2}}+xy-28$ be a map defined on ${{\mathbb{R}}^{2}}$

1) Show that , $-28\le xy\le \frac{28}{3}$, when ever $f\left( x,y \right)=0$

2) Find all integer roots for $f\left( x,y \right)=0$

Solution:  1) We have $f\left( x,y \right)={{x}^{2}}+{{y}^{2}}+xy-28={{x}^{2}}+{{y}^{2}}+2xy-2xy+xy-28$

So $f\left( x,y \right)={{\left( x+y \right)}^{2}}-xy-28$  but $f\left( x,y \right)=0\Leftrightarrow {{\left( x+y \right)}^{2}}-xy=28$

We know that from the AM-GM inequality that $\frac{x+y}{2}\ge \sqrt{xy}\Leftrightarrow {{\left( x+y \right)}^{2}}\ge 4xy$

But ${{\left( x+y \right)}^{2}}=xy+28$ so $xy+28\ge 4xy\,\,\,\And \,\,\,{{\left( x+y \right)}^{2}}\ge 0$

Thus $28\ge 3xy\Rightarrow xy\le \frac{28}{3}$ and $xy+28\ge 0\Rightarrow xy\ge -28$ hence $-28\le xy\le \frac{28}{3}$

2 ) let $p=xy\,\,\,\And \,\,S=x+y$ be the product and sum for the above equation.

Observe that ${{S}^{2}}={{\left( x+y \right)}^{2}}=p+28$ is prefect square thus $S=\pm \sqrt{p+28}$

So we are interesting in the integer solutions only for $S$ such that $-28\le p\le \frac{28}{3}$

$p=-28\Leftrightarrow S=0$ , $p=-27\Leftrightarrow S=\pm 1$ , $p=-24\Leftrightarrow S=\pm 2$

$p=-19\Leftrightarrow S=\pm 3$ , $p=-12\Leftrightarrow S=\pm 4$ , $p=-3\Leftrightarrow S=\pm 5$

$p=8\Leftrightarrow S=\pm 6$ ,

Observe that when $p\in \left\{ -28\,\,,\,\,-27,-19\,\,,\,-3 \right\}$ there is no integral roots for $x\,\And \,y$

$p=-24\And \,\,S=2,-2\Leftrightarrow \left\{ \left( -4,6 \right),\left( 6,-4 \right),\left( -6,4 \right),\left( 4,-6 \right) \right\}$

$p=-12\,\,\And S=4,-4\Leftrightarrow \left\{ \left( -2,6 \right),\left( 6,-2 \right),\left( -6,2 \right),\left( 2,-6 \right) \right\}$

$p=8\,\,\And \,\,S=6,-6\Leftrightarrow \left\{ \left( 2,4 \right),\left( 4,2 \right),\left( -4,-2 \right),\left( -2,-4 \right) \right\}$

Hence the integral roots for $f\left( x,y \right)=0$

$\left( x,y \right)\in \left\{ \begin{align}
  & \left( -4,6 \right),\left( 6,-4 \right),\left( -6,4 \right),\left( 4,-6 \right),\left( -2,6 \right),\left( 6,-2 \right) \\
 & ,\left( -6,2 \right),\left( 2,-6 \right),\left( 2,4 \right),\left( 4,2 \right),\left( -4,-2 \right),\left( -2,-4 \right), \\
\end{align} \right\}$