Factorisation of a given polynomial into another with integer roots Source: Challenge and Thrill of Pre-College Mathematics
"Find all integers $a$ such that
$$(x-a)(x-12)+2$$
can be factored into $(x-b)(x-c)$, such that $b$ and $c$ are integers."
My attempt: Simplifying the given polynomial and by the given condition, we get
$$x^2+(-12-a)x+(12a+2)=x^2-(b+c)x+ bc,$$ from where we obtain the system
$$b+c= (12+a)$$
$$bc= 12a+2$$
I can't proceed with the same process. A hint would be appreciated.
 A: Hint: Using comment by MEEL, we have:
$(b-a)(b-12)=-2$
Following cases can be considered:
1): $b-12=-1\Rightarrow b=11\Rightarrow 11-a=2\Rightarrow a=9$
Similarly:
2): $b-12=-2 \Rightarrow b=10\Rightarrow 10-a=1\Rightarrow a=9$.
3): $b-12=2$, $b-a=1$
4): $b-12=1$, $b-a=-2$
These two cases give $a=15$
A: We could also look at the properties of the family of "upward-opening" parabolas described by $ \ y \ = \ (x-a)·(x-12) + 2 \ \ . $  Each member of this set must pass through the point $ \ (12 \ , \ 2 ) \ \ , $ which we shall see imposes an important restriction on its zeroes $ \ b \ $ and $ \ c \ \ . $  This forces the family of parabolas to be arranged symmetrically about the line $ \ x = 12 \ \ . $
The vertex of a particular parabola is then located at $ \ x = \frac{12 + a}{2} \ = \ \frac{b + c}{2} \ \ $  (which is the geometric interpretation of the relevant Viete relation).  The additional requirement that the zeroes be integers will prove to be a very stringent constraint as we then must have for these zeroes $ \ b \ , \ c \ \le \ 11 \ $ or $ \ b \ , \ c \ \ge \ 13 \ \ . $
The difference of the zeroes is given by $ \ b - c \ = \ \sqrt{\Delta} \ \ , $ which must therefore be an integer.  We can then say that either the larger of the two zeroes must satisfy $  \ \frac{12 + a}{2} + \frac{\sqrt{\Delta}}{2} \ \le \ 11 \ \Rightarrow \ a + \sqrt{\Delta} \ \le \ 10 \ \  $ or the smaller of the zeroes must satisfy $  \ \frac{12 + a}{2} - \frac{\sqrt{\Delta}}{2} \ \ge \ 13 \ \Rightarrow \ a - \sqrt{\Delta} \ \ge \ 14 \ \ . $
As for the discriminant of $ \ x^2 - (12 + a) + (12a + 2) \ $ itself, we have $$ \ \Delta \ \ = \ \ (12 + a)^2 - 4·1·(12a + 2) \ \ = \ \ a^2 + 24a + 144 - 48a - 8 \ \ = \ \ (a - 12)^2 - 8 \ \ . $$  Our earlier stated specification means that this number must be a perfect square.  What we also learn from this is that since we must have $ \ \Delta \ \ge \ 0 \ $ in order for a particular parabola to have real zeroes, then either $ \ a - 12 \ \ge \ \sqrt8 \ \Rightarrow \ a \ \ge \ 12 + \sqrt8 \ \approx \ 14.83 \ \ $ or $ \ -(a - 12) \ \le \ \sqrt8 \ \Rightarrow \ a \ \le \ 12 - \sqrt8 \ \approx \ 9.17 \ \ . $  The integral condition calls for these inequalities to be interpreted as $ \ a \ \ge \ 15 \ \ $ or $ \ a \ \le \ 9 \ \ . $
While we could devote effort to further analysis from here, we quickly see that our inequalities have very narrowly "pinned-down" the permissible values of $ \ a \ \ : $
$$ \mathbf{a = 9} \ \ \Rightarrow \ \Delta \ = \ (9 - 12)^2 - 8 \ = \ 1 \ \ \Rightarrow \ \ a + \sqrt{\Delta} \ = \ 10 \  ,  \ 12 + a \ = \ 21  \ \  \Rightarrow \ \ b \ , \ c \ \ = \ \ 10  \ , \ 11 $$
and the associated polynomial is $ \ x^2 \ - \ 21x \ + \ 110 \ \ ; $
$$ a = 8 \ \ \Rightarrow \ \Delta \ = \ (8 - 12)^2 - 8 \ = \ 8 \ \ \Rightarrow \ \ a + \sqrt{\Delta} \ = \ 8 + \sqrt8 \ > \ 10 \ \ \text{[inadmissible]} \ \ ;  $$
$$ \mathbf{a = 15} \ \ \Rightarrow \ \Delta \ = \ (15 - 12)^2 - 8 \ = \ 1 \ \ \Rightarrow \ \ a - \sqrt{\Delta} \ = \ 14 \  ,  \ 12 + a \ = \ 27  $$ $$  \Rightarrow \ \ b \ , \ c \ \ = \ \ 13  \ , \ 14 $$
and the associated polynomial is $ \ x^2 \ - \ 27x \ + \ 182 \ \ ; $
$$a = 16 \ \ \Rightarrow \ \Delta \ = \ (16 - 12)^2 - 8 \ = \ 8 \ \ \Rightarrow \ \ a - \sqrt{\Delta} \ = \ 16 - \sqrt8 \ < \ 14 \ \ \text{[inadmissible]} \ \ .  $$
Consequently, there are only two solutions, represented in the graph below by the two parabolas located symmetrically about $ \ x = 12 \ \ . $

