Find values of $a$

Question

How do I find an integer $a$ such that that :

$$ (x-a)(x-12)+2$$

can be factorized as $(x+b)(x+c)$ where $b$ and $c$ are integers too.

I have tried expanding the equation and taking the determinant as $k^2$ . But I can't proceed any further from that

Answer 1

The equation $$x^2-(a+12)x+12a+2=0$$ has integer roots.

For which we need that there is natural $k$ for which $$(a+12)^2-48a-8=k^2$$ or $$(a-12)^2-8=k^2$$ or $$(a-12-k)(a-12+k)=8$$ and solve a number of systems.

I got $a=9$ or $a=15$.

Answer 2

Alternatively, equate them: $$(x-a)(x-12)+2=(x+b)(x+c) \Rightarrow \\ x^2-(a+12)x+12a+2=x^2+(b+c)x+bc \Rightarrow \\ \begin{cases}-a-12=b+c\\ 12a+2=bc\end{cases} \stackrel{12R_1+R_2}{\Rightarrow} \\ -142=12b+12c+bc \Rightarrow \\ (b+12)(c+12)=2 \Rightarrow (b,c)=(-10,-11);(-11,-10);(-13;-14);(-14;-13) \Rightarrow \\ a=-b-c-12=9;15.$$