7. Find the positive integer $x$ that makes $x^{2}-60$ a perfect square

Find the positive integer $x$ that makes $x^{2}-60$ a perfect square
7. Solution: Let $x^{2}-60=y^{2}$, then
$$(x-y)(x+y)=60$$

If $x \neq y(\bmod 2)$, then $x+y=x-y+2 y \equiv x-y \neq 0(\bmod 2)$, so both $x-y$ and $x+y$ are odd, which contradicts equation (6). Therefore, $x$ and $y$ must both be odd or both be even, thus $2|(x-y), 2|(x+y)$. Since $60=4 \times 15$, there are only the following four cases:
( 1$)\left\{\begin{array}{l}x-y=2, \\ x+y=30\end{array}\right.$
(2) $\left\{\begin{array}{l}x-y=6, \\ x+y=10,\end{array}\right.$
(3) $\left\{\begin{array}{l}x-y=10, \\ x+y=6,\end{array}\right.$
(4) $\left\{\begin{array}{l}x-y=30, \\ x+y=2\end{array}\right.$

The solutions are
$$\left\{\begin{array}{l}
x_{1}=16 \\
y_{1}=14
\end{array},\left\{\begin{array} { l } 
{ x _ { 2 } = 8 , } \\
{ y _ { 2 } = 2 , }
\end{array} \left\{\begin{array} { l } 
{ x _ { 3 } = 8 , } \\
{ y _ { 3 } = - 2 , }
\end{array} \left\{\begin{array}{l}
x_{4}=16 \\
y_{4}=-14
\end{array}\right.\right.\right.\right.$$

Thus, the integers $x$ that make $x^{2}-60$ a perfect square are 16 or 8.