Factorisation of a given polynomial into another with integer roots

Question

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.

Answer 1

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 $a=15$

Answer 2

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 \ \ . $