Example 1 A positive integer, when added to 100, becomes a perfect square. If 168 is added to it, it becomes another perfect square. Find this number.


Let the number to be found be $x$. By the problem, there exist positive integers $y, z$, such that
$$\left\{\begin{array}{l}
x+100=y^{2} \\
x+168=z^{2}
\end{array}\right.$$

Eliminating $x$ from the above two equations, we get
$$z^{2}-y^{2}=68$$

Factoring the left side of this binary quadratic equation and standardizing the right side, we have
$$(z-y)(z+y)=2^{2} \times 17$$

Since $z-y$ and $z+y$ are both positive integers, and $z-y < z+y$, it follows from (1) and the unique factorization theorem (Unit 3 (5)) that we must have
$$\left\{\begin{array} { l } 
{ z - y = 1 , } \\
{ z + y = 2 ^ { 2 } \times 1 7 ; }
\end{array} \left\{\begin{array}{l}
z-y=2, \\
z+y=2 \times 17 ;
\end{array} ;\left\{\begin{array}{l}
z-y=2^{2}, \\
z+y=17
\end{array}\right.\right.\right.$$

Solving these systems of linear equations one by one, we find $y=16, z=18$, hence $x=156$.