# Arithmetic progression and polynomials

Suppose the quadratic polynomial $$p(x)=ax^2 + bx + c$$ has positive coefficients $a, b, c$ such that these are in AP in the given order. If $m$ and $n$ are the integer zeros of the polynomial then what is the value of $m+n+mn$? I have tried quadratic formula but ain't getting the answer.

Since $m$ and $n$ are roots of the polynomial we have \begin{align*} ax^2+bx+c&=a\left(x-m\right)\left(x-n\right)\\ &=ax^2-a(m+n)x+amn \end{align*} Then, $$m+n=-\frac ba\quad\text{and}\quad mn=\frac ca$$ So, since $a,b$ and $c$ are in AP we get $$m+n+mn=-\frac ba+\frac ca=\frac{c-b}a=\frac{b+b-a-b}a=\frac{b-a}a=\frac ba-1$$ and $\frac ba =-(m+n)$, then $$m+n+mn=-(m+n)-1\quad \implies \quad mn+2(m+n)+4=3$$ So $$(m+2)(n+2)=3$$ Now, since $3$ is a prime we have $(m,n)\in\left\{(-1,1),(1,-1),(-3,-5),(-5,-3)\right\}$. Notice that $(m,n)=(-1,1)$ or $(m,n)=(1,-1)$ implies $b=0$, but we know that $a,\,b$ and $c$ are positive. So, $(m,n)=(-3,-5)$ or $(-5,-3)$, and then $$\boxed{m+n+mn=-8+15=7}$$
• Arunabh I don't take the common difference as $b$. I put $$c=b+\underbrace{b-a}_{\text{common diff.}}$$ – Ángel Mario Gallegos Feb 8 '18 at 6:20