65. Find all integers $a$ such that the quadratic trinomial

$$
y=(x-a)(x-6)+1
$$

can be represented as the product $(x+b)(x+c)$, where $b$ and $c$ are integers.
$\triangle$ Let's substitute $x=-b$ into the identity

$$
(x-a)(x-6)+1=(x+b)(x+c)
$$

We get:

$$
(-b-a)(-b-6)+1=0, \quad(b+a)(b+6)=-1
$$

From this, either $b+a=1, b+6=-1$, i.e., $b=-7, a=8$, or $b+a=-1$, $b+6=1$, i.e., $b=-5, a=4$.

The problem could also be solved in another way: in the identity written above, expand the brackets, equate the coefficients of like terms on the left and right sides, etc.

Answer: 4 or 8.