How can I solve this question on Quadratic Equations and Arithmetic Progression?

Question

Let $P (x) = ax^2 + bx + c$ where $a,b,c$ are in arithmetic progression and are positive constants. Let $P (x) = 0$ for $t$ and $r$, where $t$ and $r$ are integers. Find $t + r +tr$.

Answer is : $7$

I tried it as follows :
$t + r = -b/a$ ; $tr = c/a$ ; $t + r + tr = -b/a + c/a = (c-b)/a = (b-a)/a= b/a - 1$

Therefore,
$2t + 2r + rt = -1$

I could not figure out what to do next. Please help.

Answer 1

1. c/a = t*r is integer, b/a =-t-r is integer => we can set a = 1;

2. $x^2 + bx + 2b-1 = 0$ has 2 integer roots

So, $b^2 - 8b + 4$ is a square if integer.

$x^2 = b^2 - 8b + 4 = (b-4)^2-12$

$(b-4-x)(b-4+x)=12$ - now you can check all factorizations of 12 and find possible b and x :)

Answer 2

$$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$

$$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$

then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.

$$p(x)=ax^2+a(1+k)x+a(1+2k)$$

$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad (1)$$

So

$$k^2-6k-3=q^2 \Rightarrow (k-3)^2-12=q^2 \Rightarrow (k+q-3)(k-q-3)=12$$

and split $12$ as a product of two integer and find all possible values for $k$.

Example:

\begin{cases} k+q-3=6 \Rightarrow k+q=9 \\ k-q-3=2 \Rightarrow k-q=5 \end{cases}

Adding up both equations we get $2k=14 \Rightarrow k=7$

Backing to $(1)$ we get $(t,r)=(-3,-5)$ or $(t,r)=(-5,-3)$.